OURSE  i 
ALGEBRA 


HAWKES  -lUBY-TOUTON 


J 


GIFT  OF 


MATHEMATICAL  TEXTS 

FOR  SCHOOLS 

EDITED  BY 

PERCEY  F.  SMITH,  Ph.D. 

PROFESSOR  OF  MATHEMATICS   IN   THE  SHEFFIELD 
SCIENTIFIC   SCHOOL  OF  YALE  UNIVERSITY 


SECOND  COURSE  IN  ALGEBRA 


BY 

HERBERT  E.  HAWKES,  Ph.D. 

PROFESSOR  OF  MATHEMATICS  IX   COLUMBIA   UNIVERSITY 


WILLIAM  A.  LUBY,  A.B. 

HEAD   OF    THE   DEPARTMENT   OF   MATHEMATICS 
KAKSAS  CITY   POLYTECHNIC  INSTITUTE 

AND 


FRANK  C.  TOUTON,  A.M. 

FORMERLY  PRINCIPAL  OF   CENTRAL  HIGH   SCHOOL 
ST.  JOSEPH,  MISSOURI 


REVISED  EDITION 


GINN  AND  COMPANY 

BOSTON     •     NEW    YORK     •     CHICAGO     •     LONDON 
ATLANTA     •     DALLAS     •     COLUMBUS     •     SAN   FRANCISCO 


tt33 


ENTERED  AT   STATIONERS'  HALL 


COPYRIGHT,  1911,  1918,  BY 

HERBERT  E.  HAWKES,  WILLIAM  A.  LUBY 

AND  FRANK  C.  TOUTON 


ALL  BIGHTS  RESERVED 
921.3 


gbc   satftengum   3pre<< 

GINN  AND  COMPANY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


x^ 


PREFACE 

This  revision  of  the  "  Second  Course  in  Algebra  "  has  been 
carried  out  in  the  same  spirit  as  was  the  recent  revision  of  the 
"  First  Course  in  Algebra  "  by  the  same  authors.  The  exercises 
and  problems,  mainly  new,  have  been  carefully  graded,  and 
the  exposition  has  been  wholly  rewritten.  Some  advantageous 
changes  have  been  made  in  the  order  of  topics,  and  some 
chapters  for  which  no  well-grounded  demand  exists  have  been 
omitted.  Such  simplifications  of  exercises  and  exposition  have 
been  made  as  are  consistent  with  a  standard  course  in  third- 
semester  algebra. 

In  the  chapters  devoted  to  a  review  of  first-year  algebra  the 
fact  was  borne  constantly  in  mind  that  the  material  would  be 
handled  by  students  who  had  not  pursued  the  study  of  algebra 
during  the  preceding  year.  It  was  consequently  thought  desira^ 
ble  to  have  work  in  equations  come  much  earlier  than  before. 
The  subject  of  fractions,  the  topic  usually  most  in  need  of 
review,  has  received  full  and  careful  treatment.  Linear  sys- 
tems have  been  presented  without  the  use  of  determinants. 
Instead  of  treating  square  root,  radicals,  and  exponents  in 
one  chapter,  the  work  under  these  topics  has  been  made  more 
accessible  by  giving  a  separate  chapter  to  each.  The  needs  of 
classes,  even  under  almost  identical  conditions,  differ  widely, 
one  class  needing  more  review  on  a  certain  topic  than  does 
another.  Consequently  the  review  material  has  been  expanded 
so  as  to  afford  ample  work  for  any  class.  It  is  not  intended, 
however,  that  all  the  exercises  and  problems  should  be  solved 
by  any  one  student. 

V 

459908 


SECOND  COURSE  IN  ALGEBRA 

CHAPTER  I 

REVIEW  OF  FUNDAMENTAL  OPERATIONS 

1.  Order  of  fundamental  Operations.  The  numerical  value 
of  an  arithmetical  or  an  algebraic  expression  involving 
signs  of  addition,  subtraction,  multiplication,  and  division 
depends  on  the  order  in  which  the  indicated  operations 
are  performed.    It  is  understood  that 

In  a  series  of  operations  involving  addition^  subtraction^ 
multiplication^  and  division,  first  the  midtiplications  and' divi- 
sions shall  be  performed  in  the  order  in  which  they  occur. 
Then  the  additions  and  subtractions  shall  be  performed  in 
the  order  in  which  they  occur  or  in  any  other  order. 

Within  any  parenthesis  the  preceding  rule  applies. 

If  the  multiplication  of  two  numbers  or  number  symbols  is  indi- 
cated by  juxtaposition  without  any  symbol  of  multiplication,  it  is 
customary  to  think  of  the  multiplication  as  already  having  been 

9  a 
performed.    Thus  9  a  ^  7  a^  = 

EXERCISES 

Simplify : 

1.  6  -  2  +  8  -  7.  5.  36  --  4  .  3  -  8  +  2. 

2.  8  +  12-3-5.  6.  (19-3.5)(8-5)-(14-f-7). 

3.  8-6-4.  7.  50  -  4(20  -  4  •  2)-  3  +  2  -7. 

4.  24  -  2  -r-  3.  8.  12  -  8  -f-  2  +  10  .  3  -  6  +  8  .  2. 

9.  (28  -  14  .  24  -  8  -  3  +  10)  (30  -  3  -  5  -^  2). 

10.  (16  -  32  X  48  H-  8  -  8  -h  3)  -f-  (42  -  6  .  7  -  42  -  6)  .  6. 

1 


9 


2  •    SEGO-JSTB  COURSE  IN.  ALGEBRA 

'  '  Fmd'tiie  name^ieai  v.alue  of : 

11.  9ic-7  itx  =  3. 

12.  2x^-5x  +  3iix  =  4.. 

13.  t^-St''-^St-lift  =  2. 
•    14.  Does  4(2;:c-5)  +  15  =  3(£t;4-10)  if  cc  =  7?. 

15.  Does  (t  -\-4:)(t-{-3)-(t-\- 1)  (z^  +  2)=  42  if  ^  = 

ic    T^  3       •     2x        3  +  5^-2.^2. „ 

16.  Does — -  = — iix  =  2? 

X  —  1       x  -{-1  x^  —  1 

2.  Similar  terms.  Terms  which  are  ahke  iii  every  respect 
except  their  coefficients  are  called  similar. 

3.  Addition.  In  algebra,  addition  involves  the  uniting 
of  similar  terms  which  have  the  same  or  opposite  signs 
into  one  term.    For  this  we  have  the  following  rules: 

/.  To  add  two  or  more  positive  numbers^  find  the  arith- 
metical sum  of  their  absolute  values  and  prefix  to  "this  sum 
the  plus  sign, 

II.  To  add  two  or  more  negative  numbers^  find  the  arith- 
metical sum  of  their  absolute  values  arid  prefix  to  this  sum 
the  minus  sign. 

III.  To  add  a  positive  and  a  negative  number,  find  the  dif- 
ference of  their  absolute  values  and  prefix  to  this  difference 
the  sign  of  the  number  which  has  the  greater  absolute  value. 

Obviously  2  +  4+7=2+7+4  =  7+24-4,  etc.  Even 
if  we  have  a  series  of  positive  and  negative  numbers, 
the  order  in  which  they  occur  does  not  affect  their  sum. 
This  principle  of  addition  is  called  the  Commutative  Law 
for  Addition. 

For  the  addition  of  polynomials  we  have  the 
Rule.    Write  sitnilar  terms  iri  the  same  column. 
Find  the  algebraic  sum  of  the  terms  in  each  column  and 
write  the  results  ift  succession  with  their  proper  signs. 


EEVIEW  OF  FUNDAMENTAL  OPERATIONS       3 

4.  Subtraction.  For  the  subtraction  of  polynomials  we 
have  the 

Rule,  Write  the  subtrahend  under  the  minuend  so  that 
similar  terms  are  in  the  same  column. 

Then  change  mentally/  the  sign  of  each  term  of  the  sub- 
trahend and  apply  the  rule  for  addition  to  each  column. 

EXERCISES 
Add: 

1.  12,  -  8,  4-  4,  -  3,  and  6. 

2.  4  a,  3  a,  —  7a,  6  a,  and  —  2  a. 

3.  5  a  -  3  c  4-  6,  2  a  -  6  c  4-  11,  and  4  a  —  c  -  9. 

4.  3s-4i^  +  6,  -7!f  +  6s-8,  and  t-\-s^l^. 
b.  X  -  2  f  -  ?> z,  ^  f  -  ^x  -{-  2 z,  d.n&  4cz  -  f. 

6.  ^2  -  3  a  -f  1,  -  2  «2  _  7f^  _|-  6,  and  3  cv"  -  4  +  5  a. 

Write  so  that  x,  y,  or  t  shall  have  a  polynomial  coefficient; 

1.  ax  —  2x.  11.  at  —  aH  —  2  st. 

Solution.  ax  —  2x  =  {a  —  2) x.      12.  ay  -\- by  -{-  y. 

8.  ace  +  bx  -\-  ex.  IZ.  ^ ax  —  A:bx  -{-  ^ x. 

9.  5  at  —  Abt  —  2t.  14.  4  a;  —  abx  —  x. 
10.  4  2^  —  3  a?^  —  S2^.  15.   7x  —  3ax~  4:a^x. 

Write  the  following  so  that  the  binomial  will  have  a  binomial 
coefficient : 

16.  (a-3)x-c(a-3).  18.  S(a -{- b)- G(a -^  b). 

17.  A(a-x)-^c(-x-{-a).       19.   6a(^c  -  2  c)- S(x  -  2  r). 

20.  3a(x-l)-2b(x-l). 

21.  4:b(3x-2)~Sc(3x-2). 

22.  4  7w,(5  a  —  3  c)  —  6  /t(—  3  c  +  5  «). 

Subtract  the  second  expression  from  the  first : 

23.  6a,  4a.  25.  4x4-3,  8.T4-6. 

24.  8  a\  15  a^  26.  Tor^  -  10,  5x^-h  20. 


SECOND  COURSE  IN  ALGEBRA 


27.  5x-6,  20" +  8. 

28.  x^-5x-{-6,  2cc2_|_3^_io 

29.  a'-4:ac-3c',4:a'-^10ac-e'. 

30.  Sa--2b-6c,4:a-h6b-7c-2. 

31.  3a^-2c''-6ac,5a'  +  4:c'-Sac. 

32.  a^-Sa^c  +  6  ac\  7c^-h4:a^c-2  a\ 

S3.  X  —  Sf  -\-  z  —  4:ac  -\-  7 ax,  4:X  —  f  +  8  —  5ax  -h  9 ac. 
34.  a^—c-\-Sx  —  a^m  —  8 ac,  4 a*  +  m  —  8 x  — 10 ac  +  4 


Find  the  expression  which  added  to  the  first  will  give  the 
second: 

35.  Sx^  -  5x  -\-  2,  6x^  -llx  -^  S. 

36.  4:X^^Scx  +  c%10x^-j-Scx-9 g\ 

Find  the  expression  which  subtracted  from  the  second  will 
give  the  first : 

.37.  4.a'-2ab-\-b^,  7  a^ -lOab -h  6b\ 

38.  c'-ecx  -i-Sx%  9x^  -  lOcx  +  4  +  c^. 

39.  From  the  sum  of  2^^  —  4  #  —  9  and  3  ^^^  _  g  ^^  _|_  ^1^  ^^i^^,  ^.^^g 
sum  of  3  2^  -  6  +  4  ?J2  ^nd  5  -  8  2^2  +  4 1. 

40.  From  the  sum  oi  ax  —  ac  —  ^  ^  and  4  c^  —  3  ac  take  the 
sum  of  4 c^  —  8 ace  -|-  a^  and  Aac  -\-  Sax  —  5 c\ 

ORAL  EXERCISES 

1.  What  name  is  given  to  each  3  in  a^  -\-Sa?   Define  both. 

2.  Distinguish  between  an  exponent  and  a  power.  What  is 
the  meaning  of  4  in  cc*  ?  of  2  in  3^  ?  of  a  in  x^  ? 

3.  What  is  the  coefficient  of  x  in  3  a'^bx  ?  ofa^?  ofb? 

4.  What  is  the  coefficient  of  x  in  the  expressions  ax  -\-x? 
4:X  —  ax?  ex  —  ax  —  X?  What  is  the  meaning  of  3  in  3  ic  ?  of 
10  in  lOic?  of  a  in  OK? 


KEVIEW  OF  FUNDAMENTAL  OPERATIONS       5 

5.  What  is  meant  by  the  absolute  vahie  of  a  number? 
Illustrate. 

6.  What  is  a  literal  exponent  ?    Illustrate. 

8.  x^^x'^?    x«  -^  a;'  =  ?    a^^^+i  -^x^  =  ? 

9.  How  can  the  correctness  of  the  result  of  addition  be 
checked  ?  of  subtraction  ? 

10.  What  is  meant  by  arrangement  of  an  algebraic  expres- 
sion with  respect  to  a  certain  letter  ? 

11.  Arrange  a^  -\-  b*  —  4:  a^b  —  6  a%^  +  4  ab^  according  to  the 
descending  powers  of  b. 

12.  Arrange  t^  —  Sf^  —  5  -\-t  —  2t^  according  to  the  ascend- 
ing powers  of  t ;  according  to  the  descending  powers  of  t. 

13.  Is  arrangement  of  divisor  and  dividend  in  the  same 
order,  desirable  ?    Why  ? 

5.  Multiplication.  In  multiplying  one  term  by  another 
the  sign  of  the  product,  the  coefficient  of  the  product,  and 
the  exponent  of  any  letter  in  the  product  are  obtained  as 
follows : 

7.  7^e  sign  of  the  product  is  plus  if  the  multiplier  and  the 
multiplicand  have  like  signs,  and  minus  if  they  have  unlike 


II,  The  coefficient  of  the  product  is  the  product  of  the  coeffi- 
cients of  the  factors. 

III,  The  exponent  of  each  letter  in  the  product  is  determined 
hy  the  general  law  n°xnf>=  n«+  K 

For  the  multiplication  of  polynomials  we  have  the 

Rule.    Multiply  the  midtiplicand  by  each  term  of  the  multi- 
plier in  tu/m,  and  add  the  partial  products. 


SECOND  COURSE  m  ALGEBEA 

EXERCISES 

Perform,  the  indicated  multiplication-: 

1.  (6x^-3x-h4.)Sx.         5.  Qix^  -  2x  +  6)(7  -  3x^  -  x). 

2.  (3x-5)(4:X  +  3).         6.  (2x^  -  5x-i-3)(x^- 5x-{- 6). 

3.  (2s-St)(4:S  +  5t).       7.  (a*-}-2a''-4)(a^-2a-3). 
^.  (x^-.x-^2)(3x-4:).     8.  (t^-2t-\-6)(t*-3t''-ty 

9.  (2s^-3st-\-f-)(s^-{-5st-4:t^. 

10.  (2(4  _|_  8  4-  4  2^2)  (f^  +  8  -  4 1^). 

11.  (6^2  -  ttc  +  c')  («' +  c"  +  ac). 

12.  (a'*  -^2ab-[-  h^)  {a}  +  b-  ab). 

13.  {t^  —  t^  +  t)  (at^  +  a-\-  at). 

14.  (4  A2  _j_  6  Jc^  +  9  AA))  (4  A^  4.  6  A;^  -  9  M). 

15.  (3  c/.z^2  _  2  a?^^  4.  5  ^)  (6  a?^'  -  2  at  -  4.  af). 

16.  (2  ^2  _  3  c«  4.  4  ac8)  (2  a^  -  3  c«  -  4  ac«). 

,^    /a      2a«      «^/»      2^2      a^ 
^®*  V2~T""4A2~    3    ~  4.J' 

19.  (?^^  +  1  +  f)  {t'-\-l-  f)  (f'  +  1-0- 

20.  (x^  -\^  9f  -{-  l(y  -h  Sxij  -  4:x  -\-  12y)(x  -  3y  +  4). 

Find  the  value  of  : 

21.  3x^-^2x-\-r)  if  ic=:5. 

22.  3t^-it-\-Siit=-3. 

23.  9  -  8  jJ  +  5  ?{'  -  3  ^»  if  ^  =5  2. 

24.  2  ««  -  3  a^c  4-  4  ac2  -  3  c2  if  a  =  3  and  c  =^  --  2. 

25.  Does  15(x  -  a)-  6(x  -i-  a)=  3(5  a  -  3x)  itx  =  2a? 
^6.  Does  ax(a  -\-  3)  -\-  a(10  -  a^)^  x  +  3  if  x  =^  a  -  3? 

^7.  Does T -T  = 5 — ; ifa;  =  0?  ifx  =  -2? 


REVIEW  OF  FUNDAMENTAL  OPERATIONS       7 

6.  Division.  In  dividing  one  term  by  another  the  sign  of 
tlie  quotient,  the  coefficient  of  the  quotient,  and  the  expo- 
nent of  each  letter  in  the  quotient  are  obtained  as  follows : 

/.  The  sign  of  the  quotient  is  plus  when  the  dividend  mid  the 
divisor  have  like  signs,  and  minus  when  they  have  unlike  signs. 

II.  The  coefficient  of  the  quotient  is  obtained  hy  dividing  the 
coefficient  of  the  dividend  hy  that  of  the  divisor. 

III.  The  exponent  of  each  letter  in  the  quotient  is  determined 
hy  the  law  tv^  -i-  n^  =  n°-  *. 

The  method  of  dividing  one  polynomial  by  another  is 
stated  in  the 

Rule.  Arrange  the  dividend  and  the  divisor  according  to 
the  descending  powers  of  some  common  letter,  called  the  letter 
of  arrangement. 

Divide  the  first  term  of  the  dividend  hy  the  first  term  of  the 
divisor  and  ivrite  the  result  for  the  first  term  of  the  quotient. 

Multiply  the  entire  divisor  hy  the  first  term  of  the  quotient, 
write  the  result  under  the  dividend,  and  suhtract,  heing  careful 
to  write  the  terms  of  the  remainder  in  the  same  order  as  those 
of  the  divisor. 

Divide  the  first  term  of  the  remainder  hy  the  first  term  of 
the  divisor  to  get  the  second  term  of  the  quotient,  and  proceed 
as  before  until  there  is  no  remainder,  or  until  the  remainder  is 
of  lower  degree  in  the  letter  of  arrangement  than  the  divisor. 

EXERCISES 

Perform  the  indicated  division  : 

1.  (2cc2-5a;-f  3)--(2x-3). 

2.  (6a^2-13£c  +  6)-(3-2£c). 

3.  {^x^  -[-Bxy-2y'')^(x-\-2if). 

4.  (6a2  4-23o-!^-55?^2)-f-(3ci-5)5). 

5.  (6a«  +  6a2-28-26a)-(2c^  +  4).      .' 


8        SECOND  COURSE  IN  ALGEBRA 

6.  (6x^  -5x'  +  25cc»  -  17x^)-^(5x^  -  2a^«). 

7.  (6-«-7s^-8)^(6'2_^4  +  2s). 

8.  (x^-5  a'x  +  2  a«) -^ (o? -\- 2  ax  -  a").' 

9.  (2  ^*  -  12  jJ^  _  2  _|.  11 25  _  7  jj8^  ^  (^j^  _  3  ^  _  2  ^2>)^ 

10.  (23s2-13s^4-2s''-60-s)--(5  4-35-6-2). 

11.  (32a;*- 60- 2£c- 104x^  +  92 ic2)--(5  +  6x-4a;2). 

12.  (a?-2ah-\-h^-^x')^{^x-)rh-a). 

13.  (4^»c4-l-4c2-^»2)^(2c-^»-l). 

14.  (a;^  +  i»'  +  8ic2+8)H-(£c2-2x  +  4). 

15.  (27  a  -  18  a^'^  _  3  ^9  ^  3  ^J4^)  ^  (3  _  i^  +  ^4>^. 

16.  (3£c^  +  9cc2  +  2  -  5aj  -  8a;«  -  cc*)h-  (x  -  1)1 

18.  (4.0"  -{-  ^  -  ^ah  +  h^  -  Qb  +  12  a)^(h  -  ?>  -  2  a). 

19.  (£c«  -  4icy  +  lQy^)^(x'  +  2;rV  +  2x^/4-  4ic/  +  4/). 

20.  (x8  +  8/  +  12o-30«2/)^(a;  +  2?/  +  5). 

21.  (a?«  +  2/'  +  ^'  -  ^xyz)^{x  +  2/  +  ^)- 

7.  Parentheses.  The  removal  of  a  parenthesis  preceded 
by  a  plus  sign  is  governed  by  the 

Principle,  A  parenthesis  and  the  sign  before  it,  if  plus,  may 
be  removed  from  an  expression  without  changing  the  signs  of 
the  terms  which  were  inclosed  by  the  parenthesis. 

The  removal  of  a  parenthesis  preceded  by  a  minus  sign 
is  governed  by  the 

Principle.  A  parenthesis  and  the  sign  before  it,  if  minus, 
may  be  removed  from  an  expression,  provided  the  sign  of  each 
term  which  was  inclosed  by  the  parenthesis  be  changed. 

The  principle  just  stated  is  equivalent  to  the  principle 
which  holds  in  the  subtraction  of  one  polynomial  from 
another. 


EEVIEW  OF  FUNDAMEJSTTAL  OPERATIONS       9 

When  one  parenthesis  incloses  another,  either  the  outer 
or  the  mner  parenthesis  may  be  removed  first.  Usually  it 
is  best  to  use  the 

Rule.  Rewrite  the  expression^  omitting  the  innermost  paren- 
thesis, changing  the  signs  of  the  terms  which  it  inclosed  if  the 
sign  preceding  it  he  minus  aiid  leaving  them  unchanged  if  it 
he  plus. 

Comhine  like  terms  that  may  occur  within  the  new  inner- 
most parenthesis. 

Repeat  these  processes  until  all  the  parentheses  are  removed. 

EXERCISES 

E/emove  parentheses  and  simplify  : 

1.  a  +  {a-h)-{a-Zh).  3.   ^b -\_{a  -  h)-{c  -  a)']. 

2.  ^_(a-6')  +  (2c-3a).  4.  2c  -  2(a  -  c)-f  3(c  -  a). 
b.  x-2j^-^(x-y)-A.(2x-y). 

6.  4ic-a-f[-(3c-a')-(2a-3cc)]. 

7.  a_[_(a_3)  +  (3c-2a)-5«]+6c. 

8.  X  -  3  -  (ci  -  2  ic)  -  [2  (a  -  X  -  5)  -  3  (6  -  2  a)]. 

9.  2t^-^t-2t{^  +  t). 

10.  x'-^-(x-l){x-2). 

11.  ^x"  -  ^  a^  -{a  -  2x)(2>x  -  a). 

12.  6cc+(3c- 8x  +  2)-(c-a;-2).        '  . 

13.  6a;-[-(a-c)4-(3c-4a)]. 

14.  7c-[(3c-4)-6-(4a^-3a-c)]. 

15.  4.T. -2(£c-3)-3[x-3(4-2a;)+ 8]. 

16.  6cc-4(3-5cc)-4[2(x-4)+3(2ic-l)-(ic-7)]. 

17.  3x  -  2[1  -  3(2 £c  -  3  -  a)-  5{a  -(3cc  -  2  a)  -  4}]. 

18.  2t''-lt-(2t-l)(t-{-l). 

19.  (x  -  4>(£c  -  3)  -  (x  -  3)  (;r  +  2). 

RE 


10  SECOND  COUESE  IK  ALGEBRA 

20.  (a  +  b)a  -(a  —  b)b-\-3 ab. 

21.  3a(a  -  b)-(a  +  b)(a  -  b). 

22.  (x  -3)(x-4:)-{x-  5)  (x  +  3). 

8.  Important  special  products.  Certain  products  are  of 
frequent  occurrence.  These  forms  should  be  memorized 
so  that  one  can  write  or  state  the  result  without  the  labor 
of  actual  multiplication. 

I.  For  the  square  of  the  sum  of  two  terms  we  have 
the  formula        .    (a+bf  =  a" +  2ab+Ir'. 

This  expressed  verbally  is : 

The  square  of  the  sum  of  two  terms  is  the  square  of  the  first 
term  plus  twice  the  product  of  the  two  terms  plus  the  square 
of  the  second  terin. 

II.  For  the  square  of  the  difference  of  two  terms  we 

have  the  formula   ,        ,.„       9      «   i.  ,  1.0 
(a  —hy  =  a^  —  2ah-\-  b". 

This  expressed  verbally  is: 

The  square  of  the  difference  of  two  terms  is  the  square  of 
the  first  term,  minus  twice  the  product  of  the  two  terms,  plus 
the  square  of  the  second  term. 

III.  For  the  product  of  the  sum  and  tlie  difference  of 
two  terms  we  have  the  formula 

,  '    {a  +  b)(a-b)  =  a'-bi'. 

This  expressed  verbally  is: 

T^e  product  of  the  sum  and  the  difference  of  two  terms 
equals  the  difference  of  their  squares  taken  in  the  same  order 
as  the  difference  of  the  terms. 

IV.  For  the  product  of  two  binomials  having  a  common 
term  we  have  the  formula 

{x+a){x^-b)  =  jc^  -f  {a^b)x-\-ab. 


REVIEW  OF  FUNDAMENTAL  OPERATIONS     11 

This  expressed  verbally  is:  - 

The  product  of  two  binomials  having  a  common  term  equals 
the  square  of  the  common  term^  plus  the  algebraic  sum  of  the 
unlike  terms  multiplied  by  the  common  term^  plus  the  algebraic 
product  of  the  unlike  terms, 

V.  The  square  of  the  polynomial  a-\-b  —  c  gives  the 
formula  ^^^^_^y  =:  ^2  ^  ^  _^  c'+2ab~2ac-2hc. 

This  expressed  verbally  is: 

TJie  square  of  any  polynomial  is  equal  to  the  sum  of  the 
squares  of  each  of  the  terms  plus  twice  the  algebraic  product 
of  each  term  by  every  term  that  follows  it  in  the  polynomial, 

VI.  The  cube  of  the  binomial  a  +  b  gives  the  formula 

{a  +by  =  (^  +  Sa'b+3ab^  +  b". 
This  expressed  verbally  is: 

The  cube  of  the  sum  of  two  numbers  equals  the  cube  of  the 
first,  plus  three  times  the  square  of  the  fivst  times  the  second, 
plus  three  times  the  first  times  the  square  of  the  second,  plus 
the  cube  of  the  second. 

VII.  Similarly, 

{<i-bf  =  (^-Za^b+^ali'-l^. 

This  can  be  expressed  verbally  in  a  manner  similar  to  VI. 

ORAL  EXERCISES 

State  the  result  of  the  indicated  multiplication : 

1.  (£c  +  5)1  6.  (2  £c  -  If.  11.  {x"  -  xf. 

2.  {x  +  7)^  7.  {^x-  5)1  12.  {a"  +  3  af. 

3.  (2x  +  lf.  8.  {ax -If.  13.  {x-a){x  +  a). 

4.  (3  cc  +  2)-.  9.  {x  +  7  a)\  14.  {x  -  5)  (x  +  5). 

5.  {x^^^f.  10.  (2x-^ijf.  15.  (2x-l)(2a;+l). 


12 


SECOND  COUKSE  IN  ALGEBRA 


16.  {ax  +  3)  {ax  -  3). 

17.  (4£c-3c)(4£c  +  3c). 

18.  {ax  —  c)  {ax  -{-  c). 

19.  (4a  +  c)(c-4a). 

20.  {Za^2x){2x-2>a). 

21.  (a^  +  3)(.^  +  4). 

22.  {x  +  2){x  +  b). 

23.  {x  +  l){x  +  l). 

24.  (x-3)(x-5). 

25.  (cc  -  2)  (ic  -  7). 

26.  {x  -  3)  (a;  +  4). 

27.  (a;  H- 6)  (ic  -  7). 

28.  (£c  H-  2)  (£c  -  9). 

29.  (x  -  5)  (£c  +  10). 

30.  (o^cc  —  3)  {ax  -\-  5). 

31.  (2a;-l)(2^  +  3). 

32.  (3cc  +  l)(3a;  +  4). 

33.  (4x-2)(4ajH-3). 

34.  {a^-3a){a''^4.a). 

35.  (a  +  ^>4-c)2 

36.  {a^-G  +  xf 

37.  (c*  +  c  -  cc)*'^. 

38.  (a-c  +  ic/ 

39.  {a  +  e-\-  Vf 

40.  (a-c  +  l)2 

41.  (a  +  c  +  2)2 

42.  (a  +  c  -h  6)-^. 

43.  (2  a -I- c  + 1)2. 

44.  (a +  20- -3)2. 

45.  (a -3  a;  4- 2)2. 


46. 

47. 


48. 

49. 

50. 

51. 

52. 

53. 

54. 

55. 

56. 

57. 

58. 

59. 

60. 

61. 

62. 

63. 

64. 

65. 

66. 

67. 

68. 

69. 

70. 

71. 

72. 


(,,  _  3  ^  -  cf. 

a:^  +  b^  +  c'  +  </2 
-\-2ab  +  2ac  +  2ad 
+  2hc  +  2hd 
+  2cd. 

+  c  +  x^  df. 

-J-  ^  -j_  c  4-  xf. 
a  -\-h  -\-c  —  xy, 

-\-b-c-xy. 

-y  -c-  af. 

+  2a  +  c  +  yf. 
[3  X  -^  a  -\-  y  —  c-y. 

«  +  2  z»  +  3  c  +  dy. 

a  —  2c-\-x  —  Z  yy. 
2a  —  e  —  Sx  -\-  yy, 

a  +  cy. 

a  +  2)1 

[a-\-sy. 
a  —  xy. 

a  -  2y. 
a  +  1)=. 

1  -  ay. 
a  -  sy. 

a-{-2xy. 
3  a  -  x)\ 
?ya  +  2xy. 
'^^a-Sxy. 
a^  -  ay. 
x'  +  2  .t)«. 
[2  a2  _  ay. 


EEVIEW  OF  FUNDAMENTAL  OPERATIONS     13 

EXERCISES 

Find  the  following  products  and  expand  the  results : 

1.  [(^  +  y)+l][(^4-2/)-l]. 

2.  l(x-^a)  +3][(x  +  ^)-3]. 

3.  [(a^-r/)+3][(x-a)-3J 

4.  [(:r  +  4)4-c][(^  +  4)-4 

5.  [(2  a-b)  +  c^[(2a-b)-  c]. 

6.  lx-\-(b  +  c)-][x-(h  +  c)l 

7.  [x  +(b  -  c)Xx  -(b  -  c)^. 

8.  l3+(x-^J)X^-(x-^J)y 

9.  [10  -(a-  5)]  [10  +(a-  5)]. 

10.  [4.x  +  (2y-x)X^x-(2y-x)']. 

11.  From  each  corner  of  a  square  piece  of  tin  of  side  a  inches 
a  square  of  side  b  inches  is  cut.  By  turning  up  the  sides  an 
open  box  is  formed.  Show  that  a^  —  4:b^  is  the  area  of  the 
inside  of  the  box  in  square  inches. 

12.  Express  the  area  a^  —  4:b^  of  Exercise  11  as  the  product 
of  two  binomials. 

13.  Using  the  results  of  Exercise  12,  find  by  a  short  method 
the  area  of  the  inside  of  the  box  if  a  =  12  and  6  =  3;  if  a  =  95 
and  b  =  5. 

14.  The  dimensions  of  a  rectangular  box  are  d,  d  +  3,  and 
6^  +  6.  Express  (a)  the  sum  of  the  edges,  (b)  the  total  outer 
surface,  and  (c)  the  volume  of  the  box. 

15.  Assume  each  dimension  of  the  box  of  Exercise  14  equal 
to  n  inches  and  solve  as  before. 

16.  If  two  equal  boxes  of  dimensions  n,  n  —  4:,  and  n  +  5 
are  placed  end  to  end,  find  (a)  the  sum  of  the  outer  edges, 
(b)  the  outer  surface,  and  (c)  the  combined  volume. 

17.  The  area  of  a  circle  is  given  by  the  formula  tt)^,  in 
which  TT  =  ^-  and  r  equals  the  radius  of  the  circle.    What  is 


14  SECOND  COURSE  IN  ALGEBRA 

the  area  of  a  circular  ring  left  by  cutting  a  circle  of  radius  r 
from  the  center  of  a  circle  of  radius  R  ? 

18.  The  surface  of  a  cylinder  is  2  7rr(h  -]-r),  in  which  r 
is  the  radius  of  the  base  and  h  is  the  height  of  the  cylinder. 
Find  the  amount  of  tin  in  1250  cylindrical  cans  which  have  a 
circular  base  of  radius  2  inches  and  a  height  of  5  inches. 

19.  How  much  less  tin  is  used  in  making  a  cylindrical  can 
of  height  h  inches  and  radius  r  inches  than  in  making  one  of 
height  h  -\-  2  inches  and  radius  r  +  2  inches  ? 

20.  Would  more  tin  be  used  in  constructing  a  cylindrical 
can  of  height  h  and  radius  r  -^  2  or  of  height  A  +  2  and 
radius  r  ?   How  much  more  ? 

21.  Find  the  total  surface  of  a  cylinder  the  radius  of  whose 
base  is  r  +  8  and  whose  height  is  r  —  5. 

22.  The  formula  for  the  volume  of  a  cylinder  is  Trr^/i,  in 
which  r  equals  the  radius  of  the  circular  base  and  h  the  height. 
Find  the  number  of  gallons  contained  in  the  1250  cans  of 
Exercise  18,  given  that  1  gallon  contains  231  cubic  inches. 

23.  Find  the  volume  of  a  cylinder  the  radius -of  whose  base 
is  r  —  5  and  whose  height  is  r  -\-  5. 

24.  The  formula  for  the  surface  of  a  sphere  is  4  tt/^,  in 
which  r  is  the  radius  of  the  sphere.  Compare  the  sum  of  the 
surfaces  of  1728  balls  of  radius  one-half  inch  with  the  surface 
of  a  ball  of  radius  9  inches. 

25.  Compare  the  surface  of  a  sphere  of  radius  3  n  with  the 
surface  of  a  sphere  of  radius  n. 

26.  The  formula  for  the  volume  of  a  sphere  is  |-  tt;**.  How 
many  spherical  balls  of  radius  2  inches  can  be  made  from  a 
spherical  ball  of  lead  of  radius  12  inches  ? 

27.  Write  the  formula  for  the  volume  of  a  sphere  whose 
radius  is  r  —  6  inches. 

28.  Compare  the  volume  of  two  spheres  of  radii  r  and  2r 
respectively. 


CHAPTER  II 
LINEAR  EQUATIONS   IN  ONE  UNKNOWN 

9.  Definition  of  an  equation.  An  equation  is  a  statement 
of  the  equaKty  between  two  equal  numbers  or  number 
symbols. 

Thus  a  (a  —  2)  =  a^  —  2  a  and  a:  +  5  =  7  are  equations. 

Equations  are  of  two  kinds  —  identities  and  equations  of 
condition. 

10.  Identities.  An  arithmetical  or  an  algebraic  identity- 
is  an  equation  in  which  either  the  two  members  are  alike, 
term  for  term,  or  become  so  if  indicated  operations  be 
performed. 

Thus  15  -  8  =  6  +  1  and  (a  +  Z>)  (a  -  &)  =  a^  _  j2  ^re  identities, 
for  in  each,  if  the  indicated  operations  be  performed,  the  two  mem- 
bers become  precisely  alike. 

An  identity  involving  letters  is  true  for  any  set  of 
numerical  values  of  the  letters  in  it. 

Thus  the  identity  a(h  —  c)  =  ah  —  ac  becomes  2  (9  —  5)  =  18  —  10, 
or  8  =  8,  when,  for  example,  a  =  2,  &  =  9,  and  c  =  5. 

11.  Equation  of  condition.  An  equation  which  is  true 
only  for  certain  values  of  a  letter  in  it,  or  for  certain  sets 
of  related  values  of  two  or  more  of  its  letters,  is  an 
equation  of  condition,  or  simply  an  equation. 

Thus  2  a;  +  5  =  17  is  true  for  a:  =  6  only;  and  a:  +  2  ?/  =  10  is  true 
for  a:  =  8  and  y  =  1  and  for  many  other  pairs  of  values  for  x  and  y, 
but  it  is  not  true  for  x  =  ^  and  y  =  2  and  for  many  other  (though  not 
for  all)  pairs  of  values  for  x  and  y. 

15 


16  SECOND  COUKSE  IN  ALGEBRA 

J.2.  Satisfying  an  equation.  A  number  or  literal  expres- 
sion which,  being  substituted  for  the  unknown  letter  in 
an  equation,  changes  it  to  an  identity,  is  said  to  satisfy  the 
equation. 

Thus  X  =  Q  a  satisfies  the  equation  2x  +  5  a  =  17  a,  for  on  sub- 
stituting 6  a  for  X  we  have  2-Qa  +  5a  =  17a,  or  17  a  =  17  a,  which 
is  an  identity. 

After  the  substitution  is  made  it  is  usually  necessary  to 
simplify  each  member  before  the  identity  becomes  apparent. 

13.  Root  of  an  equation.  A  root  of  an  equation  is  any 
number  or  number  symbol  which  satisfies  the  equation. 

Thus  8  is  the  root  of  the  equation  3  x  +  2  =  26,  for  it  satisfies  the 
equation. 

14.  Axioms.  An  axiom  is  a  statement  the  truth  of  which 
is  accepted  without  proof.  Some  of  the  axioms  most  fre- 
quently used  are 

Axiom  I.  If  the  same  number  is  added  to  each  member  of 
an  equation^  the  result  is  an  equation. 

Axiom  II.  If  the  same  number  is  subtracted  from  each 
member  of  an  equation,  the  residt  is  an  equation. 

Axiom  III.  If  each  member  of  an  equation  is  multiplied  by 
the  same  number,  the  result  is  an  equation. 

Axiom  IV.  If  each  member  of  an  equation  is  divided  by 
the  same  number  (iiot  zero),  the  result  is  an  equation. 

Each  of  the  foregoing  axioms  is  used  in  the  solution  of  the 


EXAMPLE 

Solve^-i  =  x  +  7. 

Solution. -=z  X  +  7. 

(1) 

Multiplying  (1)  by  2, 

ox-l  =  2x-\-U. 

(Ax.  ITT) 
(2) 

LINEAR  EQUATIONS  IN  ONE  UNKNOWN       17 

Adding  1  to  each  member  of  (2),  (Ax.  I) 

5x  =  2x  +  15.  (3) 

Subtracting  2x  from  each  member  of  (3),  (Ax.  II) 

3  a;  =  15.  (4) 

Dividing  (4)  by  3,  (Ax.  IV) 

X  =  5.  (5) 

Check.    Substituting  5  for  x  in  (1),  we  have 
2^5  _  i  :=  5  _,_  7^  or  12  =  12. 

Since  substituting  5. for  x  satisfies  (1),  5  is  the  root  of  (1). 

15.  Transposition.  In  the  solution  of  the  foregoing  sec- 
tion by  the  appKcation  of  Axiom  I  to  (2),  the  term  —  1  is 
omitted  from  the  first  member  and  -f- 1  is  combined  with  the 
second  member.  Again,  by  applying  Axiom  II  to  (3),  the 
term  +  2  :?:;  is  omitted  from  the  second  member  and  —  2  2^  is 
combined  with  the  first  member. 

It  thus  appears  that  a  term  may  he  omitted  from  one 
member  of  an  equation.,  provided  the  same  term  with  its 
sign  changed  from  -{-  to  —  or  from  —  to  -\-  is  written  4n  or 
combined  with  the  other  member.  This  process  is  called 
transposition. 

Hereafter,  in  order  to  simplify  an  equation,  instead  of 
subtracting  a  number  from  each  member  or  adding  a 
number  to  each  member,  as  illustrated  in  the  foregoing 
example,  the  student  should  use  transposition,  since  it  is 
usually  more  rapid  and  convenient.  He  should,  however, 
always  remember  that  the  transposition  of  a  term  is  really 
the  subtraction  of  that  term  from  each  member  of  the  equatio7i. 

16.  Equivalent  equations.  Two  or  more  equations  in  one 
unknown,  even  if  of  very  different  form,  are  equivalent  if 
all  are  satisfied  by  every  value  of  the  unknown  which 
satisfies  any  one  of  them. 


18  SECOND  COURSE  IN  ALGEBRA 

Equations  (2),  (3),  (4),  and  (5)  of  section  14  are  each 
equivalent  to  equation  (1)  and  to  each  other,  for  all  are 
satisfied  by  the  same  value  of  the  unknown. 

Of  the  four  axioms  or  assumptions  of  section  14  we  shall 
make  constant  use.  If  the  ''  same  number  "  referred  to  in 
each  is  expressed  arithmetically,  the  result  is  always  an 
equation  equivalent  to  the  original  one.  Further,  if  identical 
expressions  involving  the  unknown  be  added  to  or  subtracted 
from  each  member  of  an  equation,  the  resulting  equation 
is  equivalent  to  the  first.  If,  however,  both  members  of 
an  equation  be  multiplied  by  identical  expressions  con- 
taining the  unknown,  the  resulting  equation  mag  not  be 
equivalent  to  the  original  one. 

Multiplying  each  member  of  the  equation  x  —  2  =  3  by  a;  —  1,  we 
get  x"^  —  Q  X  -\-  2  =  'd  X  —  '3,  or  x^  —  Q  X  +  0  =  0.  Now  this  last  equa- 
tion has  the  roots  1  and  5,  whereas  the  given  equation  has  the  root  5 
only.  Here  the  root  1  was  introduced  by  multiplying  the  given 
equation  by  a:  —  1.  Results  obtained  from  the  use  of  Axiom  III  with 
multipliers  which  contain  an  unknown  should  always  be  carefully 
checked.  When  a  root  is  obtained  which  does  not  satisfy  the  original 
equation,  this  root  should  be  rejected. 

The  use  of  Axiom  IV  when  the  divisor  contains  the 
unknown  may  result  in  the  loss  of  a  root  which  the  process 
of  checking  will  not  discover.  If  an  equation  is  divided 
hy  a  factor  containing  an  unknown^  this  factor  should  be  set 
equal  to  zero.  The  root  thus  obtained  is  a  root  of  the 
given  equation. 

For  example,  if  each  member  of  a:^  —  4  =  3  a;  -I-  6  is  divided  by 
a:  +  2,  the  result  is  a;  —  2  =  3,  whence  x-  =  5.  But  a;  =  —  2  satisfies 
a;2  —  4  =  3  a;  +  6.    This  root  was  lost  by  dividing  by  a;  -H  2. 

With  these  and  with  certain  other  rare  exceptions  which 
will  be  noted  later,  the  application  of  the  axioms  will 
produce  an  equation  equivalent  to  the  given  one. 


LINEAR  EQUATIONS  IN  ONE  UNKNOWN       19 

For  solving  equations  in  one  unknown  which  do  not 
involve  fractions  we  have  the 

Rule.    Free  the  equation  of  any  parentheses  it  may  contain. 

Transpose  and  solve  for  the  unknown  involved. 

Reject  all  values  for  the  unknown  which  do  not  satisfy  the 
original  equation. 

Checking  the  solution  of  an  equation  is  often  called  testing 
or  verifying  the  result.    For  this  we  have  the 

Rule.  Substitute  the  value  of  the  unknown  obtained  from 
the  solution  in  place  of  the  letter  which  represents  the  unknown 
in  the  original  equation.  Then  simplify  each  member  of  the 
resulting  identity  until  the  two  members  are  seen  to  be  identical. 

If  the  correct  substitution  of  the  root  for  the  unknown 
does  not  transform  the  equation  into  an  identity,  an  error 
has  been  made  in  the  solution. 

EXERCISES 

Solve,  and  check  the  results  as  directed  by  the  teacher: 

1.  hx-\-\  =  2x-\-l.  3.  l  +  £c  +  5£c  +  17  =  0. 

2.  6cc  +  10  =  10x4-2.  4.  15cc  =  3(4£c-5). 

5.  2(£c  +  2)-(a;  +  5)=0. 

6.  3(2£c-l)-(5a^-l)=0. 

7.  3(2a;-7)-2(5-2^)-f-l  =  0. 

8.  3(£c4-2)-5<2x-3)-:0. 

9.  6(cc  +  4)-4(a;  +  2)- 0. 

10.  4(6j^4-2)-3(7  25  +  3)=0. 

11.  6(4z^-5)-ll(2^-3)=0. 

12.  6(2a^-l)+4(3-4a;)=0. 

13.  4(4ic-l)+3-2(34-^)=0. 

14.  5  7i-9(2  7i-f  4)-2(7z,-9)=0. 

15.  4(x-2)+3(2-x)-3*  =  6(ir  +  l).     • 


20  SECOND  COUESE  IN  ALGEBRA 

16.  Sn  -  5(4:  -  71)=  5  -  S(l-\-  n). 

17.  (x-\-l)(x-2)=x^-[-3. 

18.  (x  -{-5)(x  +  l)  =  (x-  3) (x-2)-^  10. 

19.  (4cc-3)(2x-5)-(4a;-7)(2x-l)=0. 

20.  (x  +  2)2  -f-  48  =  (x  -  4)2. 

21.  (a^  +  3)2  +  40=(cc  +  5)2. 

22.  (2x-\-Sy  =  4.(l-xy. 

23.  (x  +  4)2  -  (2  -  £c)2  =  84. 

24.  (x-^  3)  (6  £c  4-  5)  -  (2  a:  +  4)  (3  £c  -  8)  =  38. 

25.  (1  -  x)  (x  +  2)  +  (x  +  3)  (x  +  4)  =  0. 

26.  (2/_4)(6-2/)  +  (2/  +  2)(2/-4)=0. 

27.  (x  +  4)(x4-3)  =  (x  +  2)(x  +  l)+42. 

28.  (2v-S)(3v  +  2)-(4.-6v)(l-v)=0. 

29.  (5x-3)(4-6x)  +  (3x  +  4)(10x-21)-9  =  0. 

30.  3(x  +  2)(x-4)+5(cc-l)(x  +  3)  = 

4(2x-l)(.T-2)+l. 

31.  X  —  2a  =  4:a  —  X. 

32.  c  —  ic  =  X  —  c. 

33.  8s  —  x  =  X  —  4r. 

34.  ax  —  2  ah  =  A:  ah  —  ax. 

35.  2t  ax  —  5  ac  =  2f  ac  —  ax. 

36.  3c(ic- 2a)=2c(a-ic). 

37.  m  (x  —  5  tt)  —  3  m  (^i  —  ic)  =  0. 

38.  x(h-\-a)=  ah-{-a^.     • 


39.  x(a  —  g)=  a 


2  6'2 


4c2. 

41.  2ax-4a2-f  4a  =  l4-a^. 

42.  ax-a2  +  5a  =  e-j-3x. 


LINEAE  EQUATIONS  IN  ONE  UNKNOWN      21 

43.  ax  +  1  —  a^  =  x. 

44.  2mx-\-5m-3  =  3x-\-2m^. 

45.  (x  -j-  a)(x  -{-  b)=  x"  -\-  2a^  -^  Sab, 
,  46.  (a  —  c){x  —  m)  =  (m  —  c)  (x  —  a). 

47.  J  ax  -  Slab  =  l^a?  +  lbh^  -  bhx. 

48.  a^x  —  a^-\-Sax  =  S  —  10a-{-x. 

49.  c(l  +  a:)+m(ic  +  l)-ic(m4-c  +  l)=0. 

50.  m  (m  —  2  x)  +  2  am  =  a  (2  £c  —  a). 

17.  Solution  of  problems.  In  the  solution  of  problems  lead- 
ing to  simple  equations  the  following  steps  are  necessary : 

/.  Read  the  problem  carefully  aiid  find  the  facts  which  will 
later  be  expressed  by  the  equation. 

II.  Represent  the  unknown  number  by  a  letter  and  express 
any  other  unknown  involved  in  terms  of  this  letter. 

III.  Express  the  conditions  stated  in  the  problem  as  an 
equation  involving  this  letter. 

IV.  Solve  the  equation. 

V.  Check  by  substituting  in  the  problem  the  value  found 
for  the  unknown. 

In  the  preceding  sentence  the  words  "  in  the  problem  " 
are  of  importance,  for  substituting  the  value  found  m  the 
equation  would  not  detect  any  errors  made  in  translating 
the  words  of  the  problem  into  the  equation. 

PROBLEMS 

1.  The  sum  of  two  consecutive  numbers  is  975.  Find  the 
numbers. 

2.  One  number  is  five  times  another,  and  their  sum  is  102. 
What  are  the  numbers  ? 

3.  The  sum  of  two  numbers  is  54.  Twice  one  of  them 
equals  ten  times  the  other.    What  are  the  numbers  ? 


22       SECOND  COURSE  IN  ALGEBRA 

4.  The  sum  of  three  consecutive  even  numbers  is  1044 
Find  the  numbers. 

5.  The  product  of  two  consecutive  even  numbers  is  1416 
less  than  the  product  of  the  next  two  consecutive  even  num- 
bers.   Find  the  numbers. 

6.  Of  four  consecutive  numbers  the  product  of  the  sec- 
ond and  fourth  exceeds  the  product  of  the  first  and  third  by 
201.    Find  the  numbers. 

7.  One  pupil  is  four  years  older  than  another.  Eight  years 
ago  the  first  was  twice  as  old  as  the  second.  Find  their  ages  now. 

8.  Two  men  are  48  and  18  years  of  age  respectively.  How 
many  years  hence  will  the  older  be  twice  as  old  as  the  younger  ? 

9.  One  man  is  three  times  as  old  as  another.  Fifteen  years 
ago  the  first  was  six  times  as  old  as  the  second.  Find  their 
ages  now. 

10.  A's  age  is  double  B's.  Twelve  years  ago  B's  age  was 
one  fourth  of  A's.   How  old  is  each  ? 

11.  A  is  twice  as  old  as  B,  and  C  is  three  times  as  old  as  D. 
B  is  8  years  older  than  D.  In  ten  years  the  sum  of  their  ages 
will  be  113.    How  old  is  each  ? 

12.  If  each  side  of  a  square  is  increased  7  feet,  its  area  will 
be  increased  329  square  feet.    Find  the  side  of  the  square. 

13.  A  certain  rectangle  is  8  feet  longer  than  it  is  broad.  If 
it  were  2  feet  shorter  and  5  feet  broader,  its  area  would  be 
60  square  feet  greater.    What  are  its  length  and  breadth? 

14.  A  certain,  rectangular  plot  of  ground  is  15  yards  longer 
than  it  is  wide.  If  it  were  20  yards  shorter,  it  would  have  to 
be  40  yards  wider  in  order  to  have  the  same  area.  What  are 
its  dimensions  ? 

15.  A  certain  square  plot  is  surrounded  by  a  border  6  feet 
wide.  The  area  of  this  border  is  816  square  feet.  What  is  the 
side  of  the  square  ? 


LINEAR  EQUATIONS  IN  ONE  UNKNOWN       23 

16.  A  certain  picture  is  4  inches  longer  than  it  is  wide,  and 
the  frame  is  2  inches  wide.  The  area  of  the  framed  picture 
is  192  square  inches  greater  than  that  of  the  picture  alone. 
What  are  the  dimensions  of  the  picture  ? 

17.  Can  a  rectangle  of  perimeter  112  inches  be  drawn  which 
has  a  length  5  inches  greater  than  twice  the  width?  If  so, 
give  its  dimensions. 

18.  A  sum  of  |15.75  consists  of  dollars,  quarters,  and  dimes. 
If  there  are  6  more  quarters  than  dollars,  and  twice  as  many 
dimes  as  quarters,  find  the  number  of  coins  of  each  kind. 

19.  A  certain  sum  consisting  of  quarters,  dimes,  and  pennies 
amounts  to  $8.62.  The  number  of  dimes  equals  twice  the  num- 
ber of  quarters,  while  the  number  of  dimes  and  quarters 
together  is  2  greater  than  the  number  of  pennies.  Find  the 
number  of  coins  of  each  kind. 

20.  A  collection  of  109  coins  is  made  up  of  quarters,  dimes, 
and  nickels.  There  are  7  fewer  dimes  than  quarters,  and  3 
less  than  five  times  as  many  nickels  as  dimes.  Find  the 
amount  of  the  collection. 

21.  The  sum  of  the  digits  of  a  certain  two-digit  number 
is  14.  If  the  order  of  the  digits  is  reversed,  the  number  is 
decreased  by  36.    Find  the  number. 

22.  Change  the  word  "decreased"  to  "increased"  in  Prob- 
lem 21  and  solve. 

23.  The  digits  of  a  certain  three-digit  number  are  consecutive 
odd  numbers.    If  the  sum  of  the  digits  is  15,  find  the  number. 

Facts  from  Geometry.*  The  area  of  a  circle  is  the  square  of 
the  radius  multiplied  by  7r(7r  =  ^Y^-  approximately).  This  is 
expressed  by  the  formula  A  —  irR^. 

The  circumference  of  a  circle  equals  the  diameter  times  ir. 
The  usual  formula  is  C  =  2  ttR. 

24.  If  the  radius  of  a  given  circle  is  increased  14  inches,  the 
area  is  increased  1232  square  inches.    Find  the  first  radius. 


24  .SECOND  COURSE  IN  ALGEBRA 

25.  By  adding  2  inches  to  the  radius  of  a  circle  whose 
radius  is  8  inches,  how  much  is  the  circumference  increased? 
the  area  ? 

26.  Substitute  R  inches  for  8  inches  in  Problem  25  and 
solve.    Interpret  your  results. 

27.  Imagine  that  a  circular  hoop  1  foot  longer  than  the  cir- 
cumference of  the  earth  is  placed  about  the  earth  so  that  it 
is  everywhere  equidistant  from  the  equator  and  lies  in  its  plane. 
How  far  from  the  equator  will  the  hoop  be  ? 

28.  Compare  the  result  of  Problem  21  with  the  one  obtained 
when  a  similar  process  is  carried  out  with  a  sphere  18  inches 
in  diameter,  instead  of  with  the  earth. 

29.  If  the  height  of  a  square  is  increased  4  feet  and  its 
length  is  increased  twice  that  amount,  the  area  of  the  figure 
will  be  increased  248  square  feet.  Find  the  side  of  the  square 
and  the  area  of  the  rectangle. 

30.  The  rope  attached  to  the  top  of  a  flagpole  is  5  feet 
longer  than  the  pole.  The  lower  end  of  the  rope  just  reaches 
the  ground  when  taken  to  a  point  25  feet  from  the  base  of  the 
pole.    Find  the  height  of  the  pole. 

31.  The  length  of  a  given  rectangle  is  three  times  its  width. 
A  second  rectangle  is  9  inches  shorter  and  1  inch  wider  than 
the  first,  and  has  a  perimeter  one  half  as  great.  Find  the 
dimensions  of  each  rectangle. 


CHAPTER  III 
FACTORING 

18.  Definition  of  factoring.  Factoring  is  the  process  of 
finding  the  two  or  more  algebraic  expressions  whose 
product  is  equal  to  a  given  expression. 

In  multiplication  we  have  two  factors  given  and  are  required  to 
find  their  product.  In  division  we  have  the  product  and  one  factor 
given  and  are  required  to  find  the  other  factor.  In  factoring,  how- 
ever, the  problem  is  a  little  more  diJfficult,  for  we  have  only  the 
product  given,  and  our  experience  in  multiplication  and  division  is 
called  upon  to  enable  us  to  determine  the  factors. 

19.  Rational  expressions.  A  rational  algebraic  expression 
is  one  which  can  be  written  without  the  use  of  indicated 
roots  of  the  letters  involved. 

Thus  2,  ox,  3?/— v2,  and  a^  are  rational  expressions.  In  this 
chapter  factors  which  involve  radicals  will  not  be  sought. 

20.  Integral  expressions.  If  a  rational  expression  can 
be  written  so  as  not  to  mvolve  an  mdicated  division  in 
which  an  unknown  letter  occurs  in  a  denominator,  it  is 
said  to  be  integral. 

Thus  3,  7  a,  ->  and  4  a;  —  3  are  integral  expressions.   In  this  chap- 
o  1 

ter  factors  which  involve  fractions  will  not  be  sought. 

21.  Prime  factors.  An  integral  expression  is  prime  when 
it  is  the  product  of  no  two  rational  integral  expressions 
except  itself  and  1. 

«^  25 


26  SECOND  COURSE  IN  ALGEBEA 

It  must  be  remembered  that  to  factor  an  integral  expres- 
sion means  to  resolve  it  into  its  prime  factors. 

The  methods  of  this  chapter  enable  one  to  factor  integral  rational 
expressions  in  one  letter  which  are  not  prime,  as  well  as  some  of 
the  simpler  expressions  in  two  letters.  No  attempt  is  made  even  to 
define  what  is  meant  by  prime  factors  of  expressions  which  are  not 
rational  and  integral. 

There  is  no  simple  operation  the  performance  of  wliich 
makes  us  sure  that  we  have  found  the  prime  factors  of  a 
given  expression.  Only  insight  and  experience  enable  us 
to  find  prime  factors  with  certainty. 

A  partial  check  that  may  be  applied  to  all  the  exer- 
cises in  factoring  consists  in  actually  multiplying  together 
the  factors  that  have  been  found.  If  the  result  is  the 
original  expression,  correct  factors  have  been  found,  though 
they  may  not  be  prime  factors. 

22.  Polynomials  with  a  common  monomial  factor.  The 
type  form  is  ab+ac-ad. 

Factoring,      ab  -\-  ac  —  ad  =  a(lb  +  c  —  tl), 

ORAL  EXERCISES 

Factor : 

1.  3a +  6.  10.  bax  —  2ax^. 

2.  5£c  +  15.  11.  2c  +  4c--2cc?. 

3.  a^+a.  12.  4.a-10a^-2a\ 

4.  2c-Qc\  13.  6ac-3^c4-3c. 

5.  9x^-3cc.  14.  10cc+15a;«-5«?/. 

6.  cd^c^d\  ,  15.  14a*-7««  +  Ta*-7rt'. 

7.  ax"  -  A.  16.  3  c^  -  6  c^  -h  9  c»  -  15  c\ 

8.  4fx-8c2.  17.  6rV-3r8«2  +  3/'V-3rV. 

9.  14  A  -  21  A^A;.  18.  10  xY  -  2  a-/  -f  2  a^/  -  2  xy. 


FACTORING  27 

23.  Polynomials  which  may  be  factored  by  grouping 
terms  and  taking  out  a  common  binomial  factor.  The 
type  form  is  ax+ay+bx^hy. 

Factoring, 

ax -{- ay  -{- hx -\- hy  =  (ax  +  ay')  +  (hx  +  hy) 
=  a(x-\-y')-]-b(x-^y) 
=  (_x  +  y)(a-hh'). 

EXERCISES 

Separate  into  polynomial  factors  : 

1.  2(a-\-b)-{-x(a-['b).  5.   a(c  -  d) -b(c  -  d). 

2.  S(b-^5)-{-a(b-j-5).  6.  2(x  -  y)- x{x  -  y). 

3.  5x(a  —  cy-^y(a  —  c).  7.  h(m +Sn)  — 2k(m -{-3n). 
^.2a(x-2y)-\-b(x-2y).      8.  2r(5x-4:2j)-9s(5x-4.tjy 

9.  -2a(Sh-k)—b(Sh-k). 

10.  x(r  —  s)-\- y(s  —  r). 

Hint.    Write  in  the  form  x{r  —  s)—  y{r  —  s). 

11.  2a(c-Sd)-\-b'(3d-c). 

.  12.  5r(Sm-2n)-2s(2n-Sm). 

13.  ac  -{-  ad  -}-  be  -\-  bd. 

14.  ac  -\-  2  ex  -\- 3  ay  -{-  ^  xy. 

15.  mx  —  my  -\-  nx  —  ny. 

16.  cd-3cf-\-2d-^f. 

17.  «./z,r  -|-  aki'  —  ahs  —  (7A;s. 

18.  r^s  +  2rs-  37^t-^ rt. 

19.  4  /^m  +  8  Ati  -  6  A^m  -  12  A;/i. 

20.  2  a^  —  2  ax  —  ac  +  ca?. 

21.  x^-^xy^-^x^y  +  'dy^. 

22.  aa?  +  ay  -\- bx  -{- by  -\-  ex  -\-  cy. 

23.  mr  —  2  r  +  ms  —  2  s  +  m^  —  2  ^. 


28  SECOND  COURSE  IN  ALGEBRA 

24.  Trinomials  which   are   perfect   squares.     The   type 
^^^^  ^^  a^±2ab+lf. 

Factoring,  a^±  2ab -{-b'^=(a  ±  by. 

ORAL  EXERCISES 

Separate  into  binomial  factors  : 

1.  a^-\-2ax-^x\  7.  r^  -  10  r*- +  25  si 

2.  x^-2xt/-\-f.  8.   9  +  6a-\-a\ 

3.  m^  -  2  7nn  +  n^  9.  16  -  8  ao;  +  aV. 

4.  a;2_|_4^_^4  IQ,  9(r2-12x?/4-4/. 

5.  4_4a;H-ic2  11^  1  4- 10 a^>  +  25 a^^A 

6.  a^-4: ah  +  4 Z-'l  12.  {a  +  5)^.-  2(a  +  6)  +  1. 

13.  (r-.s)2-6(r-.9)+9. 

14.  4  (a  +  2)2  -  12  c  {a  +  2)  +  9  <?. 

15.  9  +  6(a  +  x)  +  (a4-ic)l 

16.  16-8(a-2a:)4-(^-2ic)'l 

17.  (^  +  ^»)2  +  4  (^  -h  />)  (c  4-  2)  +  4  (6'  +  2)1 
18."  {a  +  Z;)2  -  6(a  +  ^)  (c  -  ^)  4-  9(c  -  6/)l 

19.  a2n_i2a^^  +  36. 

20.  x?""  -  14  x^/*  +  49  ?/^ 

25.  A  binomial  the  difference  of  two  squares.     The  type 
form  is  cfi  —  W 

Factoring,  cfi—l)^={a-\-  b)  (a  —  b). 

More  generally, 

«2^2aft4-^2_^4-2cc?-c?2 

=  a^+2ah  +  b^-((^-2cd+  d^) 

=  (a-hby-(c-dY 

=  (a-}-b-\'C-d){a-^b-c-\-d). 


FACTORING 


29 


ORAL  EXERCISES 


Factor : 

1.  a^-o'^ 

10.  25^2 -36  5V. 

18. 

x^-1. 

2.  m^  -  'n?. 

11.  36a2^»2_49c2c^l 

19. 

x'-l.    ' 

3.  a^-4. 

12.  a^-25  6^ 

20. 

x''  -  81. 

4.  x"  -  9. 

13.  xy-9. 

21. 

16  -  a\ 

5.  16 -xl 

14.  xy-64^^ 

22. 

625  -  x\ 

6.  a^-16h\ 

15.  81a2_l00^»V. 

23. 

81  -  c\ 

7.   l-25cl 

16.  a* -16. 

24. 

^2m yin 

8.  9a;y-l. 

Hint.    Find  three  factors. 

25. 

^2m  _  j2n^ 

9.  160^2-25 

2/^. 

17.   a^'-h'. 

26. 

C''*  -  d""^. 

EXERCISES 

Factor : 

1.  a'-x\  7.  a?-{b  +  cf. 

2.  x'if-z\  8.  4r'^-(r-s)*. 

3.  (tt-2f-c2.  9.  9m2- 4(71 +  3)1 

4.  4 (a^  +  3)^  -  2/1  10.  (a  -  c)^  -  (cZ  +  ef. 

b.  16{x-yf-z\  II.  a (/•  +  2 s)^  -  e^ (a^  -  2/)^ 

6.   9(x-^yf-16z\  12.  x2(2^-A:)2-cc2(77^-27^)*. 

13.  4  a{a  -  cc)^  -  9  a(c  -  2  d)\ 

14.  a2cc(cR-2/)2-Z'2£c(3c-^)2. 

15.  m2  +  2m7i  +  7i2-(£c2-2x?/  +  ^). 


16.  x'-2xij  +  2f 


2  ah-  y. 


17.  cc^  +  6a3  +  9  -  ^2  4.  2  ^/^  -  z^. 

18.  a^  -  22  a  4- 121  -  &2  +  20  &c  -  100 cl 

19.  2-8£c2  +  8.x^-22/'-8?/^- 8.-;l 

20.  «2^<2  +  2a&  -f  1  -  c^  +  lOccZ  -  25 (Z'^. 

21.  ^c^-a^-2ab-l}'. 

22.  50  6^%2  _  8  ^,2  _^  8  Z>c  -  2  c^. 


80  SECOND  COURSE  IN  ALGEBEA 

23.  49  x'-  49  2/2_  14  ^_  1.  28.  r"  +  rs  -  {i^  -  s^. 

24.  121  ic»- 1-18 1/^-81 2/^.  29.  nv"  -  n- -  m  -  n. 

25.  (^2  -  ^2  -  (a  -  6).             '  30.  m-{-n-  m^  +  nl 
*26.  cc^  -  /  4-  («^a;  +  a^/).  31.  x^-4:f-\-x-2  y. 
27.  r'-rs-(7^-s').  32.  7^  -  r  -  S s  -  9 s\ 

26.  The  quadratic  trinomial.    The  type  form  is 

x^  +  bx+c. 

Since       (x  -{-  h)  {x -\- k^  =  x'^  +  (li  +  ¥)x-\-  hk, 

it  follows  that  a^  -\-hx-{-  c  can  be  factored  into  the  bino- 
mial factors  (x  -\-}i)(x-\-  k^  if  two  numbers  h  and  k  can 
be  found  which  have  the  sum  h  and  the  product  c.  The 
method  of  determining  these  factors  is  illustrated  in  the 

EXAMPLE 

Factor  x^  -2x-15. 

Solution.  Here  —  15  =  1  •  —  15  or  —  1  •  15  or  +  5  •  —  3  or  +  3  •  —  5. 
Of  these  pairs  of  factors  of  —  15  only  the  pair  —  5  and  +  3  give  the 
sum  —  2. 

Hence  z^  -  2x -15  =  (x  -  d)(x  +  3). 

For  factoring  expressions  of  the  type  x^  -{-hx  +  e  we  have 
the 

Rule.  Find  two  numbers  whose  algebraic  product  is  +  c  and 
whose  algebraic  sum  is  +  b. 

Write  for  the  factors  two  binomials  both  of  which  have  x 
for  their  first  terms  and  these  numbers  for  the  second  terms. 


EXERCISES 

Separate  into  binomial  factors  : 

1.  x^-h5x-\-  6.         4.  a''  +  8  ^  +  12. 

7.  d^-Sd-10. 

2.  x'  +  7x  +  12.       5.  d""  +  11  r/  +  18. 

8.  ^2_2^_35. 

3.  x'  +  7a;  +  6.         6.  a^ -\- 10 a -\- 25 

9.  0^-4:0-12. 

.    FACTOEING  31 

10.  r"  -r-  90.  18.  a" -^  9  a  -  10. 

11.  m^  -3m-  18.  19.  (a  -{- bf  -  2  (a -{- h)  -  8. 

12.  a^2_2ic-24.  20.  (x-yy  +  4.(x-y)-{-3. 

13.  9-lOic  +  a^l  21.  a^n  _|_  12  ct^  +  35. 

14.  7-2  -  4rs  +  35^.  22.  c^^  -  7  c«  -  18. 

15.  m^  -  7  mn  +  10  Til  23.  m'"'  -  11  m^  -  12. 

16.  1  -  5x  +  6x\  24.  a^^  -  5  a^^  -  6. 

17.  1  _|_  2 Tfc  -  24 Til  25.  yy  -2h^y  -  35. 

27.  The  general  quadratic  trinomial.    The  type  form  is 

For  many  trinomials  of  this  type  two  binomial  factors 
of  -the  form  Qix  +  It)  (rnx  +  n)  may  be  found.  The  method 
of  factormg  such  trinomials  is  illustrated  in  the 

EXAMPLE 

Factor  2  x^  +  7  a:;  — 15.  ?  a;  +  ?                (\\ 

Solution.   2a:2+7x-15  =  (?a:  +  ?)(?a:  +  ?).  ?a:  +  ?                  (2) 

To  find  the  proper  factors  we  must  supply  ^x   ■{■  ,  x 

such  numbers  for  the  interrogation  points  in  +  ?a:  —15 

(1)  and  (2)  as  will  give  2  2'^  +  7  a:  -  15     (3) 

2  3^  for  the.  product  of  the  first  two  terms  of  the  binomials, 
—  15  for  the  product  of  the  last  two  terms  of  the  binomials, 
-\-  1 X  for  the  sum  of  the  cross  products. 

Now        2x^  =  2x-x,  (4) 

and  -  15  =  -  1  •  +  15  ;  1  •  -  15  ;   +  3  .  -  5  ;   -  3  •  +  5.         (5) 

The  factors  of  2  and  — 15  from  (4)  and  (5)  may  be  substituted 
for  the  interrogation  points  in  (1)  and  (2)  to  form  the  following 
pairs  of  binomials,  each  having  a  product  containing  the  first  and 
last  terms  of  the  trinomial : 

2x-l      2x+15    2i;+l      2x-15    2a:  +  3     2ar-5     2x-Z     2a:  +  5 
a:+15       x—1         a;  — 15       x-\-l         x~5       x  +  3       x  +  6       x  —  d 


32  SECOND  COUESE  IN  ALGEBRA 

By  trial  we  find  that  only  the  seventh  pair  has  +7 x  for  the  sum 
of  its  cross  products,  which  gives  the  middle  term  of  the  trinomial. 

Therefore  2  a;^  +  7  a;  -  15  =  (2  a;  -  3)  (a;  +  5). 

After  a  little  practice  it  will  usually  be  found  unnecessary  to 
write  down  all  of  the  pairs  of  binomials  that  do  not  produce  the 
required  product. 

If  none  of  the  pairs  gives  the  required  product,  the 
given  trinomial  is  prime. 

If  an  expression  of  the  form  ax^  +  bx+c  is  not  prime, 
it  can  be  factored  by  applying  the 

Rule.    Find  two  binomials,  such  that 

I.   The  product  of  the  first  terms  is  ax^ ; 

//.    The  product  of  the  last  terms  is  +  c\  * 

///.   The  sum  of  the  cross  products  is  +  bx. 

EXERCISES 
Factor : 

1.  2oc^  +  ^x  +  2.        4.  4^2  4- 7a +  3.  7.  4;r"  +  8.T  +  3. 

2.  2a;2  +  7a^  +  6.         5.  3 .t^  +  13 .r  +  12.        8.  6 ^-^  +  7  r -}- 2. 

3.  2a2  +  9a  +  10.      6.  ^x'  +  llx  +  W.        9.  2b^-5h^2. 
10.  ?>x?-^x-\-^.  21.  9r7.2  +  3a-2. 

ir.  6(-2^7c4-2.  22.  12  7'2  +  10r-12. 

12.  3  .r^  -  11  a:  +  6.  23.  10  /-^^  -  19  r.s-  -  15  s\ 

13.  3.x' -  11  .T  4- 8.  24.  iSa'  -lla'^h-ld^bK 

14.  4  .r'  -  13  a;  +  10.  25.  6  x»  +  10  xhj  -  4 xf. 
15.10x^-29x4-10.  •          26.  2x2_5^y_3y2 

16.  12x''-llxy  +  2f.  21.-2x^^Bxy  -12y\ 

17.  2a'^  +  ^a-2.  28.  ^a^-2a%-% ah\ 

18.  S7^  +  r-2.  29.  6  a^""  -  7  a""  +  2. 

19.  2a^-a-15.  30.  3 a1»»  -  10 a"  -  8. 

20.  8 .s-2  -6s -9.  31.  10 a»"  -  a,''  -  3. 


FACTORING  33 

28.  Expressions  reducible  to  the  difference  of  two  squares. 
The  type  form  is         ^4  _|_  ^^^2^^  ^  ^^ 

If  k  has  siach  a  value  that  the  trinomial  is  not  a  perfect 
square,  a  trinomial  of  this  type  can  often  be  written  as  the 
difference  of  two  squares.  Thus,  if  ^  =  1,  the  addition  and 
subtraction  of  aW  accomplishes  this  result. 

EXAMPLES 

1.  Factor  a"  +  (1%''  -\-h\ 

Solution.     «4  +  0.%"-  +  &*  =  a^  +  2  a^U^  +  h^  -  a^l)^ 
=  (a2  +  V^y  -  (aby 
=  (a2  +  &2  4.  ab)  (a2  +  &2  _  ab). 

2.  Factor  49  h""  +  34  h^k^  +  25  k\ 

Solution.    If  36  A^p  is  added,  the  expression  becomes  a  perfect 
trinomial  square.    Adding  and  subtracting  36  h^ki^,  we  have 
49  h^  +  34  AU-2  +  25  ^4  =  49  ¥  +  70  /j^F  +  25  yl'*  -  36  ^2^^2 
=  (7h''  +  Dky  -(6hky 
=  (7A2  +  5  ^'2  4.  6  7<^-)  (7A2  +  5  ^2  _  g  J^J.y 

EXERCISES 
Factor : 

1.  x^  4-  xy  -f  ?/.  11.  25  X*  -  19  £c2  4-  9. 

2.  c*  +  c2(^2  +  d\  12.  9  ax«  -  28  axY  +  4  r/7/l 

3.  a' -\- a:'b' -\- b\  13.  4  A  +  3  c^^^^x  +  9  ^>«ic. 

4.  a''  +  3a%^-i-4.b\  14.  4a*  +  l. 

5.  m*  +  m^  +  1.  Hint.    4a*  +  1  = 

6.  x^^5x'^9.  4a*  +  4a2  +  l_4a2. 

7.  c*H- 4c2  +  J6.  15.  c^  +  4i7^ 

8.  25a*-19a2  +  i.  le.  64aV  +  icl 

9.  25cc*-lla^2  4-l.  17.  4a*^  +  ^^^ 
10.  47'*-44r^52_j_49^.4  ^^  x""^ -\- 4.y^'\ 


34  SECOND  COUESE  IN  ALGEBKA 

29.  A  binomial  the  sum  or  the  difference  of  two  cubes. 

The  type  form  is  c^zkW 

a^  +  h^  divided  hj  a-\-h  gives  the  quotient  a?' —  ah  -{-  b% 
and  a^  —  h^  divided  hj  a—  b  gives  the  quotient  a?  -\-  ah-\-  b\ 

Therefore  a^  +  5^  =  (a  +  5)  (^2  -  ab -^  b^),  (1) 

and  a^-b^  =  (a-b}(a^-i-ab-hl^').  (2) 

Formulas  (1)  and  (2)  above  may  be  apphed  as  in  the 

EXAMPLES 
1.  Factor  (^^  +  27. 

Solution,    a^  +  27  =  a^  +  S^  =  (a  +  3)  (a-  -  a  •  8  +  B^)     ' 
=  (a  +  3)  (a2  -  3  a  +  0). 


Solution.      8  -  2:8  = 

:  23  -  a:8  =  (2  -  a-)  (2^  ■ 

f  2x  +  a:2) 

=  (2  -  a;) 

(4  +  2x4-2:2). 

EXERCISES 

Factor : 

1.   x'  +  7,A 

9.  x'  -  2^ 

17.   ««+(/>2/. 

2.  6t^4-^^'. 

10.  a^-27. 

18.   c«  +  ^«. 

3.  a^4-2«. 

11.  m^  -  64. 

19.  m«  -  nl 

4.  c'  +  51 

12.  m8-(2  7i)». 

20.  (2ay-(3by. 

5.  c^'^  +  S. 

13.  8;r«-7/«. 

21.  8a:«-27v/«. 

6.  6/«  +  27. 

14.  7^-27s\ 

22.  125x«  +  8//. 

7.  a-«  +  125. 

15.  64-a;».- 

23.  (a -{- by  +  c^ 

8.  x'  -  y\ 

16.   125 -a;«. 

24.   ^?8  4- />»  +  or. -h  ^>. 

^5.  x^-y''-\-x 

-?/• 

28. 

,3_8.s.8  4.r-25. 

26.  m^  —  n^  —  m  -f-  w. 

29. 

^8»  ^  /,8n^ 

27.  ic»  -  8  /  + 

^;  —  2  ?/. 

30. 

^8m  _  pn^ 

FACTOEIKG  35 

30.  The  Remainder  Theorem.  If  any  rational  integral 
expression  in  x  be  divided  by  x  —  n,  the  remainder  is  the 
same  as  the  original  expression  with  n  substituted  for  x. 
This  fact  is  illustrated  in  the 

EXAMPLE 

Divide  x^  —  5x  -\-  6  hj  x  —  n. 
•      Solution.  x^  —   5  X  +  6 


X  +  (n  —  5^ 


(n-5)x  +  6 

(n  —  o)x  ~  n^  -{-  5  n 

n^  —  5  n  +  6  =  Remainder 

Here  the  remainder  n^—3n  +  6is  the  same  as  a:^  —  5  .r  +  6, 
the  given  expression,  when  n  is  substituted  for  x. 

EXERCISES 

1.  Divide  x^  -\-  bx  -\-  c  hj  x  —  n  and  show  that  the  remain- 
der is  n^  -{-  bn  -{-  c. 

Z.  Divide  x'^  -{-  bx  -{-  c  hj  x  —  a  and  find  the  remainder. 

3.  Divide  x^-j-  ax^-\-  bx  +  chj  x  —  n  and  find  the,  remainder. 

4.  Ill  (x* -{- x'^  —  5  x  +  S) -i- (x  —  2)  find  the  remainder  (a)  by 
division,  (b)  by  the  Remainder  Theorem. 

Solution  (b).   2^  +  2^  -  5  •  2  +  3  =  5. 

5.  In  (x^  —  X  -\-  5) -^  (^x  —  3)  find  the  remainder  (a)  by  divi- 
sion, (b)  by  the  Remainder  Theorem. 

By  use  of  the  Remainder  Theorem  find  the  remainders  in 
the  following : 

6.  (x^  -\-  x^  -  5x  -\-  S)-ir(x  -  3). 

.    7.  (x^-3x-15)-^(x-^4.). 

8.  (x^-2x''-100)^(x-5). 

9.  (x^  -  2x^  -2x-  3)^(x  -  3). 
10.  (x^-2x^-^x-  2)^(x  -  2). 


36  SECOND  COURSE  IN  ALGEBEA 

31 .  Factor  Theorem.  By  substituting  2ioTxmx^—5x-\-6 
we  obtain  4  — 10  +  6,  or  0.  Hence  a;  —  2  is  an  exact  divisor 
(or  factor)  ot  a^  —  3  x  -\-  6.  Again,  if  3  is  substituted  for  x 
m  x^  —  5  X  -\-  6,  the  expression  equals  zero.  Hence  a;  —  3  is 
a  factor  oi  x^  —  3  x  -\-  6.  These  examples  illustrate  the 

Theorem.  If  any  rational  integral  expression  in  x  becomes 
zero  when  a  number  n  is  substituted  for  x^  then  x—n  is  a 
factor  of  the  expression. 

The  Factor  Theorem  may  be  used  to  factor  some  of  the 
preceding  exercises  and,  in  addition,  many  others  which  are 
very  difficult  to  factor  by  previous  methods. 

Note.  By  means  of  the  Factor  Theorem  we  are  able  to  solve 
cubic  and  higher  equations  when  the  roots  are  integers.  The  solu- 
tion of  the  general  cubic  equation  is  one  of  the  famous  problems  of 
mathematics  and  one  which  is  accompanied  by  many  interesting 
applications.  This  problem  was  first  solved  by  the  Italian,  Tartaglia, 
about  1530,  but  was  published  by, Cardan,  to  whom  Tartaglia  ex- 
plained his  solution  on  the  pledge  that  he  would  not  divulge  it.  For 
many  years  the  credit  for  the  discovery  was  given  to  Cardan,  and  to 
this  day  it  is  usually  called  Cardan's  Solution. 

When  searching  for  the  values  of  x  which  will  make  an 
expression  zero,  only  integral  divisors  of  the  last  term  of  the 
expression  (arranged  according  to  the  descending  powers 
of  x')  need  be  tried,  for  the  last  term  of  the  factor  must 
be  an  integral  divisor  of  the  last  term  of  the  expression. 

EXAMPLE 

Factor  cc»  +  2cc-3. 

Solution.  If  X  —  n  is  a  factor  oi  x^  ■\-  2x  —  3,  then  n  must  be  an 
integral  divisor  of  3.  Now  the  factors  of  —  3  are  1,  —  1,  3,  and  —  3. 
If  1  is  put  for  X,  then  x^  -I-  2  x  —  3  equals  zero,  hence  a:  —  1  is  a  factor 
of  a:^  +  2  a:  —  3.  Dividing  a,*^  -f-  2  a;  —  3  by  a;  —  1,  we  obtain  the  quotient 
a;^  +  a:  +  3.  Since  a;^  -f-  a:  +  8  is  prime,  the  factors  of  x*  -H  2  a;  —  3 
are  a:  —  1   and  ar^  +  a:  +  3. 


FACTORING  37 

ORAL  EXERCISES 

1.  Is  X  —  1  a  factor  of  ic^  +  3  cc  —  4  ? 

2.  Is  X  -  2  a  factor  of  2x^-^x'  -  20? 

3.  Is  a  -  2  a  factor  of  a^  -  3  (^  +  2  ? 

4.  Is  X  —  1  a  factor  of  cc^  +  3  x^  —  4  ? 

5.  Is  r  4- 1  a  factor  of  r'^  —  4  r^  —  4  r  + 1  ? 

6.  Is  r  —  3  a  factor  of  2  r^  —  7^  +  5  ? 

7.  Is  5  +1  a  factor  of  3s^  -  5^^  +  8  ? 

8.  Is  A:  -  3  a  factor  of  2  A:^  -  5  ^b^  -  9  ? 

EXERCISES 

Factor : 

1.  x^-{-x-2.  8.  y^-^f  -^y-{-  9. 

2.  x^+-2xH-3.  '    9.  .T'^-Tx'-ex. 

3.  tt«4.«2_36.  10.  x^ -7x^4- 4x4-12. 

4.  ^3  ^  .^  _io.  11.  2x^  -  2x2  -  X  -  6. 

5.  ^^  +  ^"-12.  12.  x3-x2-4. 

6.  x^-2x2-5x  +  6.  13.  3x^-2x2  + 2x- 3. 

7.  ^8_x24-4x-4.  14.  a^-Q>a^  +lla'-^a. 

32.  The  sum  or  difference  of  two  like  powers.  The  type 
form  is  a"  zfc  6". 

The  cases  m  which  a^  ±  If'  is  divisible  by  a  4-  ^  or  a  —  h 
can  be  determined  by  the  Factor  Theorem. 

Thus  in  a^  —  If^  n  being  either  an  odd  or  an  even  in-" 
teger,  substitute  h  for  a.  Then  a^  —  h^  becomes  b'^  —  h^  =  0. 
Therefore  a—  b  is  always  a  factor  of  a^  —  h^. 

In  a"  —  6^,  n  being  even,  put  —  b  for  a.  Then  d!^  —  b^ 
becomes  b^  —  b'^  =  0,  since  (—  5)^  is  positive  when  n  is  even. 
Therefore  when  n  is  even  a  4-  ^  as  well  as  a  —  6  is  an  exact 
divisor  of  a^  —  b^. 


38       SECOND  COURSE  IN  ALGEBRA 

In  a'*  +  h''\  n  being  even,  put  either  -\-h  or  —h  for  a. 
Then  a^  +  V^  becomes  If^  +  h^^  which  is  not  zero.  Therefore 
^i  _|_  yti  ^g  i^eyer  divisible  hy  a-\-b  oy  a  —  h  when  n  is  even. 

In  a**  +  h''\  n  being  odd,  put  —  b  for  a.  Then  a"  +  6" 
becomes  (~  by^  +  ^'^  =  0,  since  (—  5)"  is  negative  when  ?i 
is  06?(i.    Therefore  when  n  is  odd  a-\-b  is  a  divisor  of  a^.  +  5". 

Summing  up: 

I.  a"  —  b"  is  always  divisible  by  a  —  6. 

II.  a"  —  &",  when  n  is  e^^e/i,  is  divisible  both  by  a  +  & 
and  by  a  —  &. 

III.  a"  +  &"  is  never  divisible  by  a—b. 

IV.  a"  +  &",  when  n  is  oc7c?,  is  divisible  by  c  +  6. 

ORAL.  EXERCISES 

I 
For  each  of  the  following,  state  a  binomial  factor : 


1.  x'-f. 

6.   x''-2f.              11.    7?^'^+2l 

16.   1-7-1 

2.  x^-5\ 

7.  o:^  -  2^.           12.  a'  +  8. 

17.   l-r» 

3.  27  -  a\ 

8.  2^cc^-7/l        13.  iK^+7/. 

18.  1  +  7-^ 

4.  cc^-2^ 

9.  a'^  -  c-".            14.  r^  +  2'. 

19.  l  +  /« 

5.  32^^-/. 

10.  a«-^A            15.  a^+32. 

20.  5«  +  l. 

21.  8^-1. 

23.  Is  10'  -f  1  divisible  by  11? 

22.  10^-1. 

24.  Is  10^-1  divisible 
EXAMPLE 

by9? 

Factor  x^  +  y^ 

Solution.   By  division, 

£! 

-i-^  =  x*-  xh  +  xhr  -  xy^  +  v^. 

Hence  a;'^  +  y*  =  (a:  +  ?y)  (.r^  —  ar*?/  +  a-V  —  2:?/'  +  y^). 

Note  that  the  signs  of  the  second  factor  are  alternately  plus  and 
minus.   Also  note  the  order  in  which  the  exponents  occur. 


FACTORING 

39 

EXERCISES 

Factor : 

1.  x^-^z\ 

8. 

a^-32af. 

18.  1  -  r^. 

2.  x''  +  1. 

9. 

(2x)^-2432/^ 

Hint.  Write  only  the 

first  five  terms  and  the 

3.  a^-^2\ 

10. 

a}  -  x\ 

last  term  of  the  poly- 

4. x'  +  32. 

11. 

1  -  r^. 

nomial  factor. 

5.   (ay  +  (hy 

'. 

19.   1  +  r^ 

6.  cc^-«^ 

12. 

x^  -  128. 

20.  x^  —  if. 

Hint.  Find  the  , 

see- 

13. 

x^'  +  y'K 

Hint.    Factor    first 

ond  factor  by  division 
and  observe  the    signs 

14. 

o}'^Z2x^K 

as  the  difference  of  two 
squares. 

of    the  terms  and 
order  in  which  the 

the 
ex- 

15. 

x'  4-  a\ 

21.  x«  -  y\ 

ponents  occur. 

16. 

\^t\ 

2^2.  a}''-}?-\ 

7.  a'  -  2K 

17. 

128:r^-M. 

23.  a"-"--}?^.. 

33.  General  directions  for  factoring.  The  following  sug- 
gestions will  prove  helpful  in  factoring : 

/.  First  look  for  a  common  monomial  factor^  and  if  there 
is  one  (other  than  i),  separate  the  expression  into  its  greatest 
monomial  factor  and  the  corresponding  polynomial  factor. 

II.  Then  from  the  form  of  the  polynomial  factor  determine 
with  which  of  the  following  types  it  should  he  classed^  and 
use  the  methods  of  factoring  applicable  to  that  type. 

1.  ax+ay+bx+by.  5.  ax'^  +  bx+c. 

2.  a'±2ab-\-1y',  6.  a^  +  kaW  +  b\ 

3.  a'-b'.  .  7.  a^zkb^. 

4.  x'  +  bx+c.  8.  fl"±6". 

///.  'Proceed  again  as  in  II  with  each  polynomial  factor 
obtained^  until  the  original  expression  has  been  separated  into 
its  prime  factors. 

IV.  If  the  preceding  steps  fail,  try  the  Factor  Theorem. 


40  SECOND  COURSE  IN  ALGEBRA 

REVIEW  EXERCISES 
Factor : 

1.  6x^  +  2x^-{-2x\  29.  x^  -  Sx^-4.x-^12. 

2.  5a^  +  2a^-15a-  6.       30.  x  -  x^  -  x^ -^  x\ 

3.  a^-\-4.ab-\-4:  h\  31.  (a  +  xf  +  10(a  +  x)-\-  25. 

4.  2cH-%cd\  32.  {x  +  rf-^{x  +  r)-l^. 

5.  3m«-3m2-18m.  33.  x^  -  Sx""  -  x -[- ^. 

6.  2^2  +  3  aa;  +  al  34.  a^  +  27  6i"'. 

7.  x'-lxhf-^-^y^'  35.  64c^8-|-2^«. 

8.  a^c  -  ac^.  36.  a^'^  -  a^»V  +  a^^c  -  a6c». 

9.  2x\j-2xy\  37.  (ic  +  2/)2- 6(a;  4-2^)^  +  9;5;2. 

10.  cc^  -  2  cc^  -  9  a;2.  38.  (a  -  xf  -  16(m  -  ti)^ 

11.  ac^2hG-ad-2hd.  39.  2 ic^ -  2 ic^  -  12 cc. 

12.  18  r^  -  24  r^s  +  8  7-/.  40.  10  a?c  -  15  aV  -  70  ac». 

13.  45xV-20x/.  41.  a* -11^2  +  1. 

14.  2  A%  4-  4  «2  _  30  ;j,3  42.  6t^  +  aVK 

15.  3a2-10a/>  +  3^;^.  43.  a^  +  ct'i^ 

16.  ic^  +  7  .T^  +  16.  44.  x^{x  -  1)2  -  .T(a;  -  1). 

17.  aH-%ad\  45.  4(a-^')2-12(a-^>)^  +  9<'2. 

18.  2  6i^  +  64  dK  46.  a^  -  10  aZ»  +  25  ^'^  _  c^. 

19.  m^n-\-mn\  47.  cc2_2ic(«_^,)_35(a-^)l 

20.  x^  -\-^x-  5.  48.  6^2  -  13a:  4-  6. 

21.  ax'-^a^Zx^-Vl.  49.  ««- 8  ar»  +  17  a;  -  10. 

22.  a%''  -  4  a2^»2  _|_  4  a6.  50.  ic*  -  ^if  +  a;»  -  a^V- 

23.  aV-c2-a2^_i.  51.  ^2_i2,r  +  36-a*. 

24.  xy  -  13  x\f  -  14  a;y.  52.  a""  -  b^  -\-(a-  bf. 

25.  27» -b't^s-Srs".  53.  c2-2c(Z  +  rf''-2(c-r/)-85. 

54.  2{a-\-by-^^e{a  +  b)-2c'. 

55.  r??/*  -  7  wV  +  71*. 

56.  a8  +  ^<8  +  3a'''^  +  3a^'^. 


26. 

x'- 

-l^x^ 

+  36. 

27. 

a^- 

-  a2/>8  - 

-  a»^»'-^ 

^^'^ 

28. 

x"^ 

-x\ 

FACTORING  41 

57.  ^2x^-xh/\  69.  9ic'-4i/2-3x-2y. 

58.  b{a-bf-a  +  b.  70.  x"  -b(2x-  5). 

59.  c2  +  4^2_^2_4^^  rj^    a^_S-7a^  +  Ua. 

60.  (a-2x}c'-\-(2x-a)d:\  72.  3a2-14a(^>-c)4-8(*-c)l 

61.  x^-20-\-x\  73.  aV4-4. 

62.  6a^-\-Sa^-3  al  74.  ccV  +  x'z^ 

63.  16a*  +  7a^  +  l.  75.  cc^  -  lOx  -  3. 

64.  ccy  -  64.  76.  x'  -  xS/  -  a-*  (cc^  -  ij"). 
6b.  x' -lSx^-21x\  77.  a«-5a*  +  4. 

66.  a' -  a^Z*^  +  ^2^  -  a^  78.  ti*  -  Qa^  -  «  +  3. 

67.  a*  +  8  fi^^  +  a'  +  2  «^>.  79.  x  -  1  +  a-^  -  x\ 

68.  a2  +  2a  +  l-Z''  +  2^»c-cl       80.  x^  +  x  -  y  -  f. 

%l.,x'  +  2xij  +  f-2^  a^  -lOa-1. 
82.  a''-b^-2ah{a'-b''). 

83.  2  ir^  +  3  x^  +  a;.  92.  7^'"^  -  2  A"^A:"  +  A:^^ 

84.  m^  -  8  m  -  7  m*.  93.  a^"^  -  ^'^^^ 

85.  3  (^^  -  9  a (2  a  -  3).  94.  x^'^  +  (r  +  s)^^''  4-  ^'s. 

86.  2x^-10x^-\-4:X.  95.  /^^-x'*^  +  hrx''  +  Ascc^  +  As. 

87.  10  a -7  a" -6  a\  96.  a*"*  +  a^'^b^''  +  &*'\ 

88.  a^-^-l-^-Sa'  +  Sa.  97.  a«"^  +  ^'^ 

89.  12x^-Sx-^x^-6x\  98.  a«"*  -  b^"". 

90.  a^x^  +  2  (^»ic  +  a«  99.  a^"*  +  ^>s^ 

91.  mr  —  ms  —  m-  +  ws.  100.  a^^  —  ^^^ 

101.  Solve  for  x,  ax  -\-bx  —  3a  =  3b. 

Solution.    Rewriting,  ax  +  bx  =  ^  a  -^  ^b.^ 

Factoring  in  each  member,  a:(a  +  &)  =  3  (a  +  6). 

Dividing  each  member  by  a  +  &,  a;  =  3. 

102.  Solve  for  x,  ex  -\-  Ux  =  (^  -\-  ed. 

103.  Solve  for  m,  wa  —  ?w^  -f  Jc  =  ai. 


42  SECOND  COURSE  IN  ALGEBRA 

104.  Solve  for  y,  5  ay  —  S  bi/  —  5  a^  -\-  S  ab  =  0. 

105.  Solve  for  z,  az  —  3  ad  =  bz  —  S  bd. 
"     106.  Solve  for  ic,  cic  —  2  dx  =  c^  —  4  c?l 

107.  Solve  for  m,  am  —  m  -\-  1  —  a^  =  0. 

108.  Solve  for  y,  by -^S  dy  -  b^  +  9d^  =  0. 

109.  Solve  for  r,  r(2a -7c)=  4.a^  -  28  ac  +  49  c^. 

110.  Solve  for  m,  m(Sa  —  c)—  9 ad  =  2 ec  —  6ae  —  Scd. 

111.  Solve  for  a;,  <xx  —  3  Z»x  —  a^  =  3  Z»^  —  4  ab. 

112.  Solve  for  s,  2  CIS  -  7  s  +  13  a  =  2  «2  -h  21. 

113.  Solve  ioTt,2te-2e-}-15-Se'-{-3t=0. 

114.  Solve  for  2/,  2/  4-  21 6^2  +  4  c^  =  7c?y  +  1. 

115.  Solve  for  x,  d(l  —  3a)-{-x-\-Sac  =  c-\~S  ax. 

116.  Solve  for  -z,  az  -\-  ae  —  2  ec  -\-2cd  =  2cz  -\-  ad. 

117.  Solve  for  2/,  a  -  3  ==  a?/  -  3 ?/  -  2(a  -  3). 

118.  Solve  for  £c,5cc-2cx-a(5-2c)  =  5a  +  5-2c-2«c. 

119.  Solve  for  x,  o?x  —  a^  +  x  —  a^  —  1  =  0. 

■  120.  Solve  for  x,  c»  -  c^x.  -  2cdx  -  Sd^  =  4:  d^x. 

121.  Solve  for  m,  m(a  -  2)  (a^  +  4)  =  a*  -  16. 

122.  Solve  for  s,  16  s  -  8  as  +  4  a^s  -  2  ah  +  a^s  -  32  =  a^ 

34.  Solution  of  equations  by  factoring.  The  methods 
of  factoring  enable  us  to  solve  many  equations  in  one 
unknown.  In  the  solution  of  equations  by  factoring  use 
is  made  of  the 

Principle.  If  the  product  of  two  or  more  factor's  is  zero^ 
one  of  the  factors  must  he  zero. 

Each  of  the  factors  may  be  zefo,  but  the  vanishing  of 
one  factor  is  sufficient  to  make  the  product  zero. 


FACTORING  48 

EXAMPLE 

Solve  the  equation  x^  —  x^  =  ^  r. 
Solution.    Transposing  6  x,  we  have 

a,.3  _  ^2  _  g  2^^  ^  0. 
Factoring,  x  (x  +  2)  (x  -  3)  =  0. 

Solving  the  equations  resulting  from  setting  each  factor  separately 
equal  to  zero,  we  have  x  =  0,  —  2,  and  +3. 

Substituting  these  values  in  the  original  equation,  we  find  that 
0,  —  2,  and  +  3  are  roots  of  the  equation. 

The  method  of  solving  the  above  example  is  stated  in  the 
Rule.    If  necessary,  rewrite  the  equation  so  that  the  second 
member  is  zero.    Then  factor  the  first  member,  set  each  factor 
which  contains  an  unknown  equal  to  zero,  and  solve  the  result- 
ing equations. 
Check  as  usual. 

In  the  above  example  the  division  of  each  member  oi  x^  —  x^  =  Qx 
by  x  gives  x^  —  x  =  6.  This  last  equation  has  only  the  roots  —  2  and 
3,  while  the  original  equation  had,  in  addition,  the  root  0. 

Again,  if  the  equation  a;^  —  5  x  +  6  =  a;  —  2  is  solved  by  the  rule,  the 
roots  are  2  and  4.  If,  however,  each  member  of  a:^  —  5  a:  +  6  =  a;  —  2 
is  divided  by  x  —  2,  the  resulting  equation  a-  —  3  =  1  has  the  root  4 
only.  In  each  of  these  cases  the  root  would  not  have  been  lost  if 
the  factor  used  as  a  divisor  had  been  set  equal  to  zero  and  the  equa- 
tion thus  obtained  had  been  solved.  The  solution  of  an  equation 
is  often  simplified  by  the  use  of  this  method. 

EXERCISES 

Solve  by  factoring : 

1.  x2-4  =  0.  5.  4x^  =  25x. 

2.  ar^-5x  =  0.  6.  y^=^A:y. 

3.  a;2  -  aa;  =  0.  7.  a;^  +  5  =  6x. 

8.  2x'  +  x  =  Q. 


44  SECOND  COUESE  IN  ALGEBRA 

9.  m:^-a7n-2a^=0.  25.   Sx^ -\-71x^  -  9x' =  0. 

10.  x^-2ax-Sa^  =  0.  26.  (2x-Sy -(5x-\- 6y  =  0. 

11.  f-y  +  2  =  2y\  27.  (x-r)2-(x-s)2=0. 

12.  £c2-9«^2_^_^3^^()_  28.  (£c-3)2-2(a;-3)=8. 

13.  ic«  -  ax^  -  12  a^ic  =  0.  29.  cc«  +  50-25x-2«2=o. 

14.  cc»-a2£c  +  2aa;2_2a8=0.  30.  2^/^  -  2^^  _  g  y  +  8  =  0. 

15.  x«  +  £c2- A-ax  =  0.  31.  x^-ax^-4:a^x-\-4:a'^=0. 

16.  4y  4- 19/ -52/^  =  0.  32.  cc^  +  8  +  6ar»  +  12ir  =  0. 

17.  /_77/-6  =  0.  33.  7/-^Sf-\-Sy-\-l  =  0. 

18.  7/^-13  7/2  4-36  =  0.  34.  acx^ -{-bcx -{- adx +bd=0. 

19.  x®-5a:«=-4x.  35.  .x^  -  7x  -  8  =  a^  4- 1. 

20.  x^-7x^  =  -6x.  36.  40^^  _^  cc -,  1  =  2a;  -  1. 

21.  3r«4-24/-2  4-48r=:0.  37.  3.^2  _  j-j^^  _j_  g  ^  3^  _  2 

22.  2;»  +  12^- 6«2_8  =  0.  38.  a^x^  -  2  ax  -  S  =  ax  -  S. 

23.  3a;^+72ic  4- 33x^  =  0.  39.  x^ -2x^ -15x  =  x^ -5x. 

24.  5.s2-f  12s-3s^  =  0.  40.  Gx^  +  7a;  -  3  =  3a;  -  1. 

35.  The  highest  common  factor.  The  highest  common  factor 
(H.C.F.)  of  two  or  more  monomials  or  polynomials  is  the 
expression  of  highest  degree,  with  the  greatest  numerical 
coeificient,  which  is  an  exact  divisor  of  each. 

Thus  the  H.C.F.  of  28  a%^  and  42  aV)''  is  14  a^^  The  H.C.F.  of 
x^  —  4:Z  and  x^  —  5  a:^  4  6  a;  is  a;  (a:  —  2),  or  a:^  —  2  x. 

EXAMPLE 

Find  the  H.  C.  F.  of  9  a;^  -  36  x^  and  3  a-'  -  12  a;«  4- 12  x*. 
Solution.    Factoring,  we  have 

9  a;4  -  36  a:2  =  S^x^  (x  +  2)  (x  -  2), 
3  r'  -  12  a:6  4  12  a:6  =  3  x^  (x  -  2)2. 
Therefore  the  H.C.F.  is  8  x'^(x  -  2),  which  equals  3  .r»  -  6  x% 


FACTORING  45 

The  method  used  in  the  preceding  solutions  for  finding 
the  H.C.F.  of  two  or  more  monomials  or  polynomials  is 
stated  in  the 

Rule.  Separate  each  expression  into  its  prime  factors.  Then 
find  the  product  of  such  factors  as  occur  in  each  expression^ 
using  each  prime  factor  the  least  number  of  times  it  occurs  in 
any  one  expression. 

If  two  or  more  polynomials  have  no  common  factor 
other  than  1,  then  1  is  their  H.C.F.,  and  the  polynomials 
are  said  to  be  prime  to  each  other. 

EXERCISES 
Find  the  H.C.F.  of  the  following: 

1.  12,  18,  24.  5.  30  c^^,  45  c^^  l^cH\ 

2.  15,  25,  40.  6.  2Sa%\  A2  ab',  70a%\ 

3.  24,  60,  72.  7.  66  c%  132  c'x^,  165  cV. 

4.  12  a^,  30  a',  36  a\  8.  a^  ^  2  ab  +  b%  a^  -  b\ 
9.  3  a^  -  3 b'',  9(a  -  bf,  3  a^  -  3  b\ 

10.  ax^  —  2  axy  +  ay^^  a^x^  —  a'^y'^,  2  ax^  —  2  ay^. 

11.  2  aV  —  2  A^,  4  am^  — 12  amn  +  8  an^,  10  am  —  10  an. 

12.  25  x^  -  25  x\  10  x^  -  20  xhj  -  30  xtf,  5x'-5  xy\ 

13.  3  X*  -  6  x%  6x^-24t  xSf,  12  x""  -  96  xhf. 

,    14.  24  a«  -  6  a%'',  48  a«  +  24  a%,  48  a''  -  48  a%  -  36  a%''. 

15.  5 x^  -  160 x^,  15 cc^  -  60 x\  25  x'  -  200a;^ 

16.  18  a*  -  2  a%%  12  a«  -  8  a'b  -  4  c^^^^^  30  a'^P  +  10  a^6l 

17.  4  X*  -  4  xy,  5  x^-  5  a^y,  8  x^i  -  8  xy. 

18.  6^5-3a*5  +  2a3^^  a^  -  2  a*6  -  a^^^2  +  2  a^^^. 

Note.  The  most  famous,  and  in  some  respects  the  most  perfect, 
treatise  on  elementary  mathematics  ever  written  is  Euclid's  "Ele- 
ments." About  one  third  of  the  material  of  the  thirteen  books  treats 


46  SECOKD  COURSE  IN  ALGEBRA 

topics  which  to-day  would  be  considered  arithmetical  in  character. 
In  appearance  and  language,  however,  they  are  all  geometrical,  for 
Euclid  represents  quantities  not  by  numerals,  as  we  do  in  arithmetic, 
or  by  letters,  as  we  do  in  algebra,  but  by  lines.  Book  VII  contains 
the  earliest  statement  of  a  general  method  for  finding  the  G.C.D.  of 
two  numbers.  This  method,  though  never  necessary  in  elementary 
mathematical  work,  is  so  perfect  and  beautiful  from  a  scientific  point 
of  view  that  until  recently  it  remained  in  elementary  treatises  on 
algebra  and  arithmetic  by  force  of  tradition.  It  is  a  great-  tribute  to 
Euclid's  genius  that  he  was  able  to  devise  so  perfect  a  method  for 
the  process  that  all  the  efforts  of  two  thousand  years  have  been 
unable  to  improve  it  essentially.  It  'is  of  fundamental  importance 
in  advanced  portions  of  algebra. 


CHAPTER  IV 
FRACTIONS 

36.  Operations  on  fractions.  The  change  of  a  fraction  to 
higher  or  to  lower  terms,  and  the  addition  and  the  sub- 
traction of  fractions  in  both  arithmetic  and  algebra,  depend 
on  the 

Principle.  Tlie  numerator  and  the  denominator  of  a  fraction 
may  he  multiplied  hy  the'  same  expression  or  divided  hy  the 
same  expression  without  changing  the  value  of  the  fraction. 


Thus 

3  3.4      12 

4  "4.4  ~  16' 

^18       18  ^  6      3 

^"^    30      30-^6      5 

Similarly, 

a      a  •  n      an 
b       b  •  n      bn 

,     a      a  ^  n      n 

and     -  = =  -. 

0       b  -^  n       b 

It  should  be  noted  that  by  the  application  of  this  principle  a 
fraction  is  changed  in  form  but  not  in  value. 


ORAL  EXERCISES 

Divide  both  numerator  and  denominator  of 


^•if^^'S.  5.gby3.^        8.^by.  +  2. 

47 


48 


SECOND  COUESE  IN  ALGEBRA 


Reduce  to  lower  terms  : 

9 


10. 


11. 


12. 


^  ax 
~Za' 

10  xy 

2{x- 


5) 


^x{x-by 


13. 


14. 


15. 


3(a--2) 
^2-4* 
5a;+10 
2£c-h4  * 

x(x  +  ^)' 


EXERCISES 
Reduce  to  lowest  terms  : 
2a«&2 


19. 


10  6^2^,8 

15  x]/^ 
24.5  x^y' 
x''-9 
2x^-{-6x' 

5x^  -\-  5xy 
25x^-25xhf' 
2a^-{-S  d'b  +  8  ah^ 
5  rt.^  -  20  (^ 
48.T^-6x/ 
32x*-32icV+8.Ty 

3a^»-375 
2ic2+10£c  +  50' 
^,8  +  6  a^^  4-  9  </^»'^ 


^a^b  +  ^ah'' 


17. 


18. 


x2-6a'  +  9 


9. 

ro. 
11. 

12. 
13. 
14. 
15. 

16. 

4a2 


16. 

1  Tl 

x' 

X^ 

'  +  4 
-16 
a;«4-8 

x^ 

+  4a;  +  4 

1  Q 

4 

-2.^  + J-- 

iC^  + 


2c^  +  2c^"-4cZ^ 

2c^-^6Acd' 
16c*-40c^^+16cV-^' 
ac?  —  3aa;  +  2cd  —  6cx 
ad  —  Sax  —  cd  -\-  3 ex 

2x^-Sx''  +  Sx  ' 
'4a;^  +  6x-40 
X  —15  a  -\-  6  ax  —  10 
5x*-4:0x 
Sx^-96  ' 
12^^+10^^^ -12 <?. 

4:-\-9e(^-12a 
Sx*y-\-SxY-\-Sy^ 
5  x^y  -\-  5  xxf  +  5  £cy 


ic2-f-4«a-4-4a2-  9 
32^^^^-.^' 


32+16icH-4£c» 

2     .    -I 


^  +  1 


21. 


(f)'+(i)'   • 


FRACTIONS  49 

37.  Changes  of  sign  in  a  fraction.  In  the  various  opera- 
tions on  fractions  three  signs  must  be  considered :  the  sign 
of  the  numerator,  the  sign  of  the  denominator,  and  the 
sign  before  the  fraction.  Since  a  fraction  is  an  indicated 
quotient,  the  law  of  signs  in  division,  when  considered  in 
connection  with  the  sign  of  a  fraction,  gives  the  following 

identity: 

+  a_      — «_      — a_      -\-a  ^ 

+  :^  -  "^  i:^  ~  ~  IjTi  ~  "  -  ^  * 

From  the  above  we  have  the 

Principle.  In  a  fraction  the  signs  of  both  numerator  and 
denominator,  or  the  sign  of  the  numerator  and  the  sign  before 
the  fraction,  or  the  sign  of  the  denominator  and  the  sign  be- 
fore the  fraction,  may  be  changed  without  alte7'ing  the  value 
of  the  fraction. 

In  applying  this  principle  to  a  polynomial  denominator 
or  numerator,  like  Sx^  —  3x  -{-  2,  we  change  its  sign  by 
changing  the  signs  of  every  term  of  the  polynomial.  If 
the  polynomial  denominator  is  in  factored  form,  like 
(a;  —  3)  (2  a?  —  1)  (3  a;  4- 1),  its  sign  is  changed  by  changing 
the  sign  of  ang  one  factor  or  of  an  odd  number  of  factors. 

ORAL  EXERCISES 

Read  the  following  fractions  in  three  additional  ways,  using 
the  above  principle : 

1.^.  2.-?-.  3.^-  4.-/^. 

X  —  X  I  —  X  \—  a 

Change  the  sign  of  the  following  indicated  products : 

5.  (£t;-3)(£c  +  3).     6.  (2x4- 5)(3a^- 5).     7.  {x-a){x-b). 

Change  in  two  ways  the  sign  of  the  following : 

%.  {x  ^-V){x  -V){x}  ^-V).        10.  (2x-3)(3a:-4)(4a^-5). 

9.  {x-a){x-b){c-x).         11.  (cc-2)2(£c-l)(.T  +  2). 


50  SECOND  COURSE  IN  ALGEBRA 

12.  Find  the  indicated  product  in  Exercise  8.  Then  change 
the  sign  of  two  factors  and  find  the  product.  Compare  the 
results. 

13.  Make  a  general  statement  of  which  the  result  of  Exer- 
cise 12  is  an  illustration. 

Change  the  form  of  the  following  fractions  so  that  each  will 
contain  the  factor  cc  —  1  in  its  denominator  : 

14.  A:-  16.  -f^-  18.+.    '" 


1—  X  1—  X  2  —  X  —  X^ 

^^'        1-x''  x(l-xy  ^  -.i^2x-x^ 

By  proper  changes  of  sign  make  the  denominators  of  the 
following  fractions  as  nearly  alike  as  possible : 

3  x-\-S 


20. 
21. 


(x  -2)(x-j-  2)       (2  -x)(2  +  x) 

a  h 


(a  —  h){c  —  a)  (b  —  c)       (b  —  a)  (c  —  a)(b  —  c) 
c 


(a  —  h){a  —  c)  (c  —  b) 

38.  Lowest  common  multiple.  The  lowest  common  multiple 
(L.C.M.)  of  two  or  more  rational  integral  expressions  is 
the  expression  of  lowest  degree,  with  the  least  numerical 
coefficient,  which  will  exactly  contain  each. 

Thus  the  L.C.M.  of  6,  8,  and  12  is  24,  and  tlie  L.*..M.  of  a-c, 
ac^,  and  2  abc^  is  2  a^bc^. 

EXAMPLE 
Eind  the  L. CM.  of  a;«  -  4 cc,  2  a-^  -  4 x^,  and  4 .r*  -  8 x^ 
Solution.   Factoring,     x^  — 4:x  =  x(x  +  2) (x—  2), 
2x^-4:X^  =  2x^(x-2), 
4x*-  8x^  =  ix'^(x-2). 
The  L.C.M.  iHix\x  +  2)(x-  2). 


FKACTIONS  51 

For  such  expressions  as  can  be  readily  factored  the 
method  of  finding  the  L.C.M.  is  stated  in  the 

Rule.  Separate  eacJi  expression  into  its  prime  factors.  Then 
find  the  product  of  all  the  different  prime  factors^  using  each 
factor  the  greatest  number  of  times  it  occurs  in  any  one 
expression. 

ORAL  EXERCISES 

Find  the  L.C.M.  of  the  following: 

1.  4,  8,  12.  5.  x-2,  x^  -  4,  2(ic  -f-'2). 

2.  25,  50,  75.  6.  x^  -  8,  2{x  -  2),  o^^  +  2a;  +  4. 

3.  4a;,  6£c2,  8:r.  7.  ^^2  _  9^  ^2  _  g^  _l_  9 

4.  2aly',  4:a%  6ab^  S.  x  -  5,  x^  -  x  -  20. 

9.  a^-4.,a^-^a-\-4.,  S(a  +  2). 

10.  a^-2a-\-l,l-a,  a. 

11.  x^  -'125,  x^-^5x  +  25,5-x. 

12.  ax(a^  -  x^),  a^(x^  -  a^),  x\a  -\- x). 

EXERCISES 

Find  the  L.  C.  M.  of  the  following : 

1.  a^-  ab^,  a^-2ab-\-  b^,  a"  +  ab. 

2.  ax  -\-  ay  -\- bx  -\-  by,  o?  —  V^,  x^  —  y'^. 

3.  c'-^cd-^  16  d"",  c"  -  16  d"",  6^  +  4  cd. 

4.  ^^  _  4^5  _  21^2^  7^  -  49 s^  r^  -  95^. 

5.  2m'  -\-mn-  10n%  4^^  -  25n^,  m«  -  Sn\ 
e.  Sr'-Sr\l-2r-^7^,r'-l. 

7.  125  -  n%  n^  -  25/1,  50n-i-  IOti^  +  2?i«. 

8.  a^  -  32,  4  -  a^,  5a^-\- 10a,  5a -10. 

9.  «'  -  Tec  +  6,  a;2  _  3^  4.  2,  ic^  -f  2ic  -  3. 

10.  x^-^4.x^-4.x-16,x^-4:,x'-\-2x-S. 

11.  a?^  —  £c,  x^  —  x'*,  x^  -^  x^  -\-  X,  x^  --  X. 

12.  4  a^  -  20  ^  +  25,  25  -  4  a^,  2  a^  -f-  15  a  +  25. 


52  SECOND  COURSE  IN  ALGEBRA 

39.  Equivalent  fractions.  Two  fractions  are  equivalent 
if  one  can  be  obtained  from  the  other  either  by  multiply- 
ing or  by  dividing  both  numerator  and  denominator  by  the 
same  expression. 

Two  fractions  having  unlike  denominators  cannot  be 
added  or  subtracted  until  they  have  been  reduced  to  re- 
spectively equivalent  fractions  having  like  denominators. 

EXAMPLE 

Reduce  to  respectively  equivalent  fractions  having  the  lowest 
common  denominator  (L.C.D.): 

2  5  .  2x 

and 


3  X      x^  —  4  x^  —  2x 

Solution.   Rewriting  with  denominators  in  factored  form,  we  have 

2  5  J  2a.- 

and 


Then 


and 


3x       (a:  +  2)(x-2)  x{x-2) 

TheL.C.D.  is  3  x(x  +  2)  (a;  -  2). 

2__   2  (a:  +  2)  (a:  -  2) 
3x~3a:(a;  +  2)(a;-2)' 
5  ^  5.3a; 

(a:  +  2)(a;-2)       3a;(a;  +  2)(a;-2)' 
2a:        _      2      _       2  •  3  a: (a;  +  2) 
a;(a;  -  2)  ~  X  -  2  ~  3  ar(a;  +  2) (a:  -  2) ' 


To  change  two  or  more  fractions  (in  their  lowest  terms) 
to  respectively  equivalent  fractions  having  the  L.C.D.  we 
have  the 

Rule.  Rewrite  the  fractions  with  their  denominators  in 
factored  form. 

Find  the  L.O.M.  of  the  denorhinators  of  the  fractions. 

Multiply  the  numerator  and  the  denoming,tor  of  each  frac- 
tion hy  those  factors  of  this  L.  C.  M.  which  are  not  found  in 
the  denominator  of  the  fraction. 


FRACTIONS  53 


EXERCISES 

Change  to  respectively  equivalent  fractions  having  the  L.  C.  D.  : 

25  ^         2x  — 1  —x-\-  5 

1.     TT-^  7^^'  5. 


3ic    2cc'  x2-6£c  +  9    -2x  +  6 

^      2x          3                   ^          cc  +  l  — a? 

2.   ;;>— ^-  6.  -7i z r^J 


a; 

— 

2' 

x  +  2 

3 

2c^ 

a^ 

— 

9' 

'a-3 

3 

a 

2  a' 

7. 


8. 


x^  —  5ic  +  6    —  x^  +  4 

2  a  '        5 «. 

2a2  +  3a  +  l'2«^-hl' 
2a       ^4-3  2^2  +  5 


(^-2)2^-2  a  +  2   3a  +  13a2  +  7a-f2 

a     2  a        5a 


9. 
10. 


2^    3x   x'-\-2x 
2x4-1  5 


a;2-2ic  +  4   a;^  +  8    x -^  2 
2x-9  -2 


11.  -^ ^» 

12 


^5_32    _a;8  +  2x2   x'-{-2x^  +  4.x^+Sx'-\-16x 
ax  —  hx  —  ar  -\-  br    x'  —  i^    o^  —  IP" 


ax  -\- hx  -\-  ar  -\-  hv    a?  —  ly^    x?  —  r^ 
2x         5x  4  3ic  +  2 

13.    -;; -y- J 


x^-S    2-x    4:-x'   x^  +  2x'  +  4.x 

40.  Addition  and  subtraction  of  fractions.  To  find  the 
algebraic  sum  of  two  or  more  fractions  in  their  lowest 
terms  we  proceed  as  in  the 

EXAMPLE 

Find  the  algebraic  sum  of  — H — r r- 

°  a        a'-       ah 

Solution.    TheL.C.D.  isa% 

Rewriting  with  common  denominators,  and  adding  numerators, 
^®  ^^^®  2^  .  ?i!  _  5l^  =  2  «&'  +  3  ?>2  -  5  g 


64       SECOND  COUKSE  IN  ALGEBRA 

Check.    Substituting  2  for  a  and  3  for  b,  we  have 
6      9      5  _  36  +  27-10 
2      4      6"  12 

53^53 
12"  12' 

For  finding  the  algebraic  sum  of  two  or  more  fractions 
we  have  the 

Rule.  Reduce  the  fractions  to  respectively  equivalent  frac- 
tions having  the  lowest  common  denominator.  Write  in  succes- 
sion over  the  lowest  common  denominator  the  numerators  of  the 
equivalent  fractions^  inclosing  each  polynomial  numerator  vn  a 
parenthesis  preceded  hy  the  sign  of  the  corresponding  fraction. 

Rewrite  the  fraction  just  obtained^  removing  the  parentheses 
in  the  numerator. 

Then  combine  like  terms  in  the  numerator  and^  if  neces- 
sary^ reduce  the  resulting  fraction  to  its  lowest  terms. 

Check.  Set  the  original  expression  equal  to  the  final  result 
and  substitute  in  each  member  numerical  values  for  the  letters 
involved.    The  equation  should  be  an  identity. 

EXERCISES 

Find  the  algebraic  sum  of : 

2x      3^_2£r  .        c  2c 


510       15  •        "c-4c-f3 

2c5c^c  a  2a         1 

^    a-b      a-Zb      Sa  3a;-l  3 


ab  a"  y  x'-2x^-Z      x-2 

2r8^-\-r      r-Qs _  s^  2g  +  l  4 

r's      ~   3r^  ~2r  x^-^       x-Z 

5a        2  ,.  «-3      a- 3         5 

10. 


3      3  a-^-4      2-«  '  a  +  2 


FEACTIOKS  55 

6^  +  2  c-5  ,,  ^  x^ 

^''-  c(c-2)      8-c^  ^-2 


13. 


^^      _  1  _  ^m_-2^      17.  a'  +  ab  +  P-  -^ 
m  —  1      m       m^  —  1  a  —  o 


14.2  +  -^-  .  ''■''-f^S-^- 

X  —  7 

20.  -i-        ^^         ^^^ 


a;  +  2     '       X  ^x'  +  2x-2 

c-^2d  1        •       2 


22. 
23. 


c^^2Gd  +  d''      d-G      c  +  2^ 
a_3  2-a  1 


a,2_2o^-3      a2_3^_^2      l-a^ 


^^*     4-^2  a-2    "^   a  +  2 

a  +  2  a-\-2 


25. 


a2_7^_,_12      6  +  ^2- 5a      6a -a^- 8 

2S.*±_«_2(*-^). 
0  —  a         \a       a  —  01 

28.   ,.     '.+  '       .+ \*'.     .      ,+  "  +  • 


(^  —  c)  (c  —  a)       ac  —  a?"  —  be  -\-  ab       (a  —  b)(b  —  c) 

x^  —  (m  —  ny      m^  —  {x  —  nY      t?  —{x  —  TYif 
(x  +  ny  —  Tf?       {x  -\-  KYif  —  n^       (rti  +  rtf  —  x^ 


i 

•1  = 

=  M- 

a 

c 

ac 

b 

d 

-u' 

5.1f  = 

-i-h'- 

=  -V-. 

n 

a 

n    a 

na 
h  * 

56  SECOND  COUESE  IN  ALGEBRA 

41.  Multiplication  of  fractions.  In  algebra,  as  in  arith- 
metic, the  product  of  two  or  more  fractions  is  the  prod- 
uct of  their  numerators  divided  by  the  product  of  their 
denominators. 

Thus 
Similarly, 
and 

In  like  manner 

Integral  and  mixed  expressions  are  reduced  to  fractional 
form  before  the  multiplication  is  performed.  Factors  com- 
mon to  any  numerator  and  any  denominator  are  canceled 
the  same  number  of  times  from  each. 

EXAMPLE 

Multiply  20«^(l-i)(^^^j^). 

Solution.    Writing  the  above  in  fractional  form,  we  have 
20  g^   g^  -  4  6  a 

1     '     a2      '5a3-10a2* 

Factoring  and  canceling, 
4 

1  ^  -^rtJ^XjOr — -^  a 

a 

To  find  the  product  of  two  or  more  fractions  or  mixed 
expressions  we  have  the 

Rule.  If  there  are  integral  or  mixed  expressions,  reduce 
them  to  fractional  form. 

Separate  each  numerator  and  each  denominator  into  its 
prime  factors. 


FRACTIONS  5T 

Cancel  the  factors  (^factor  for  factor^  common  to  any 
numerator  and  any  denominator. 

Write  the  product  of  the  factors  remaining  in  the  numerator 
over  the  product  of  the  factors  remaining  in  the  denominator. 

Check,  Set  the  given  expression  equal  to  the  final  result 
and  substitute  in  each  member  numerical  values  for  the  let- 
ters.  Simplify  each  member.    The  result  should  be  an  identity. 

42.  Division  of  fractions.  For  division  of  fractions  we 
have  the 

Rule.  Reduce  all  integral  or  mixed  expressions  to  fractional 
form. 

Then  invert  the  divisor  or  divisors  and  proceed  as  in  mul- 
tiplication of  fractions. 

Check  as  usual. 

EXERCISER 
Perform  the  indicated  operations  : 
6  a       10  xip'  ^    a^  —  4        4  a 


25  xV     4  A  2^2 

^     ^         3(^       4c  ^      3^2 

2.  5c.  TT^^-TT-,-  9. 


lOc^^    ^d  x  +  3       ^x^ 

laV'     ^^cU  jQ    a}-\  ^a?  5a-5 


^cd"-    Ua%  ■  2a       a'-2a-{-l    3(l-a) 

^     Sah    21  c^    20  a%^  ,,     (^      ,   1\/      a''     \(  1\ 


4 


^    2a^    _^  12.  A-tV 

2  '  3x''  l^a  13.  If  H-lf: 


6.  lOa.I^.^.  14.     1^^  ^^- 


21  (T^  "•  15 


ax" 


4:m\     6ns     2mV  120 e^d      100  cd 


9  7iV     4.m^s    2mn^         *"'    42c^x    *   147  c^ic^ 

RE 


58  SECOISTD  COUESE  IN  ALGEBRA 

24 a;*     _8^  ^  _48^  22  mV  .  52  m'n  ,  275  n-x 

^<^xif'  22x'y  ''  \2\xY  35^3     '    125 ar^  '    39m-^x 

_    15  rs^      300ri(«    1605^*  ,^      5x^  ^{x-yf  1 

14^2^       98^2^      28rV  x-y       2^^  x{x-y) 

x  +  S    a;^-4a;"H-4  ,  x'-2x 
x^~4:'      2x-t-6^      '  x^  +  2x' 
x^^2x  +  4l  ^    a;»-8     .  x-^  ~  2a; 
a;'-2x  +  4  '  ic2  +  2x  *    x^  +  8  * 
„     2a;^  +  9a;  +  9   ^      x^  -  9       a;'-6x  +  9 
2a;2-9cc  +  9  *  4x^-9x'    2x2- 3« 

x^  —  y^  .  i^^  +  i^y  +  ^z"*   ^'^  —  ^2/  +  3/^ 

*a7^  +  y^'  x  +  y  a;  —  2/ 

a^  +  32   ,  a«  +  2a2   af" -^  2a^  +  Aa"" -^  Sa +  16 
a" -32   '  a^-2a'  a^-2a^  +  4.a^-Sa  +  16' 


25. 
26, 


x^-Qx'  +  llx-S  ,  a;2-5a;  +  6    a;'' +  1 
a;*-l  '  x^-\-2x-^l'  x-\-l' 

2x  +  5  ^  25-4a;2  ^  2:^2  + a; -15 

^3_6^2^_18.^_27  •    a;»  +  27    '         x-3 


29. 
30 


-©)"-Ki5)'©)«(if.)e)' 

-•(— D(^)fei)• 
1  4.  «2   •  \^-^       a«  +  1/      V     ««  +  1      / 


33 
34 


FEACTIONS  59 

9 


•  '^;rri^\Sa'-{-la  +  2}v''~l~V 


35 

— : — 12" 

X-  5  + 


a?  -|-  3 

Hint.    An  expression  of  this  form  is  called  a  complex  fraction.    It  is 
simply  another  way  of  writing 

\  x  +  3/      \  x  +  3/ 

X  —  2  \^  —  y         /      \x  —  y         / 


,       3 
X-1--  2  11 


X 

X  +  1  + 


38. —  b       a-\-b       a  —  b 

x__  45_  


X 


39. 


a  -{-  b       a  —  b 
a      ,       b  a-b~a+b 

b  a  b 


^  +  1  +  ^^  46.'^^  +  ^^  «  +  ^ 


a 


.     a  1  —  a 

a  -\-  1  a 

41.  ^ 

a  1  — a  48^ 


-^2 

X 

a  — 

-  a 
T 

f  1 

1   ^^- 

a;" 

•s^ 

2«/ 


a  —  i       <x  +  ^ 


&Hc 


a  +  b       a  —  b  2  b 

42.  TT^rr—TTT:-         49. 


2  a^/>"  4-  2  ^>^  A.__2      "M—  --      i_- 

^j'.^  —  />^  2  a  \  c        a       c        I 


b^cll  - 

'    4a6 

a 

60  SECOND  COUESE  IN  ALGEBRA 

50.  If  ic  =  ■; >  y  =  — ; — )  and  z  = ->  find  the  value 

6  +  c  a  -\-  c  a  -{-  0 

n  ,    ,.;  when  a  =  x—  -  and  b  =  x  -\-  - 
a/  -^b^  X  X 


51.  What  value  has    ^  ,   ,.,;  when  a  =  x—  -  and  b  —x-V-'i 


52.  1 ^-.      53.  1 ?: —      54.  1  + 


1  + ^  1- 


1  +  ^  1-|  1  +  i 

Z  6  a 

55.  Brouncker  (1620-1684)  proved  that  ir  (the  circumference 
of  a  circle  divided  by  its  diameter)  is  four  times  the  following 
fraction : 

1 


1  + 


2  + 


2  + ?^ 


24-       '^ 


2  +  etc. 

225 


(cC)  Rewrite  the  fraction,  continuing  it  to 


2  +  etc. 

(&)  Stopping  with  2  -f-  -^^- ,  find  the  difference  in  value  between 
four  times  the  value  of  this  fraction  and  3.1416,  the  approxi- 
mate value  of  TT. 

Note.  William  Brouncker,  one  of  the  brilliant  mathematicians 
of  his  time,  was  an  intimate  friend  of  John  Wallis  (see  "  First  Course 
in  Algebra,"  Revised  Edition,  p.  48).  These  two  scientists  were 
among  the  pioneers  in  the  study  of  expressions  with  a  countless 
number  of  terms. 

The  complex  fraction  in  the  exercise,  if  continued  indefinitely 
according  to  the  law  which  its  form  suggests,  is  called  an  infinite 
continued  fraction.  Brouncker  was  the  first  to  study  the  properties 
of  such  expressions. 


FRACTIOKS  61 

43.  Equations  involving  fractions.  Equations  involving 
fractions  are  solved  as  in  the 

EXAMPLE 
Solve  '  5-«  =  ^(^.  (1) 

X  X  +1 

Solution.    The  L. CM.  of  the  denominators  is  x(x  +  1). 

(1)  .  x(x  +  1),  5a:(a;  +  1)  -  8(x  +  1)  =  5^2-  5a:.  (2) 

From  (2),  5a;2  +  5a:  -  8  a:  -  8  =  5a;2  -  5a;,  (3) 

or  ■  .         2a:  =  8. 

Whence  a:  =  4. 

Check.    Substituting  4  for  x  in  (1),  5  -  2  =  3,  or  3  =  3. 

For  solving  equations  in  one  unknown  which  may  or  may 
not  ipivolve  simple  fractions  we  have  the 

Rule.  Where  polynomial  denominators  occur^  factor  them 
if  possible. 

Find  the  L.Q.M.  of  the  denominators  of  the  fractions  and 
multiply  each  fraction  and  each  integral  term  of  the  equation 
by  it,  using  cancellation  wherever  possible. 

Transpose  and  solve  as  usual. 

Reject  all  values  for  the  unknown  which  do  not  satisfy  the 
original  equation. 

EXERCISES 

Solve  the  following  for  x  and  check  as  directed  by  the  teacher : 


1. 

2       2      2^- 

2. 

¥-l  =  f  +  3. 

3. 

2x  ,  ,             7 

4. 

llx      2 
9        3  =  "- 

3x      16 

2         3 

25 
=  x--^ 

^x      21 
6         5 

^x-2 
10 

2(^-4) 
3 

3 

2(0.-2) 

2  +  5a; 
9 

62  SECOND  COURSE  IN  ALGEBRA 

^    2x  +  l      lzL?^_Qi  1^    a:  +  5_10 

^'  ~4  3~~  ~  ^^-  ^^=^  ~  3  ' 

3         7        4'  14.        ^       =  — ^ 

4  a?      16      3  a; 

1,    2x-7      3 

•  3a;      5x      45  3x-2      3 
12.  3(a;  +  2)-4(2a;-3)  +  2  =  0.  a;  +  7    "^  5 

17.  (a;  +  5)(a;  +  l)-(ir-3)(x-2)  =  10. 

1ft    _A_   .   __J___o  99    ^  +  l_x  +  4 

^**  2a;-5^2a;-l"^'  ^'^*  a;  +  3~a;  +  2* 

2a;-5^3a;-14  a;+3        3a; +  4 

*  2a;4-7~  3a;-2*  2aj -f  1  ~  6a;  -  2* 

„^    5a;  +  3       5a;  +  7  ^^     17-3a;       13-3a; 

20.   = •  24.   = 

2a;-3  2a;  3»-13       3a;-8 

«^11  —  ^/v  «,.^  —  3h  5 

21. 7i 5  =  0.  25.    — ;  =  1 


a?6  —  a;*  a;4-7  x  -\- 1 

26. 


3(1- 8a;)       2(1+ 8a;)  _   1- 34a; 
5x  8a;  —1  5x 


27._1-  +  ^^±1==1_      ^^ 


28. 
29. 


2x+l       l-2a;  4a;2-l 

c  -\-  X  _  x(c  —1) 
~  x-\-l  "  c  (a;  + 1) ' 

l+2a;  _  3  a; +1  _  8  + 3a; -4a;'' 
2a;-l        x+1    ~    2a;2  +  a;-l 


30. a^-  —  h. 

a  0 

x^a      x-\rh  x      --       x        ^ 

31. :; — ==  0  —  a.  33.  tH —:r=^Za, 


a  b  '  a  —1      a  +1 

a       0       X  ^  <(  -\-  2 


o«    1   ,   1       1                                 oA        ^          a; +16       . 
32.  -  +  -  =  -•  34.  — —rr ?r  =  4a. 


36. 


FRACTIONS  63 

2a;+l      3x-7  _9-Sx-a^ 

'x-{-32~x~x'^x-6^ 
X  -\-  cd      X  —  cd  _2  chl  —  2  cd'^ 
c  -\~  d         c  —  d  c^'—  d^ 

37.  f^-|^^  =  0. 
Lx  —  a       ox  —  c 

5x-\-  A      1.3  a; -.05      30.35 -8  a; 
^^*  '~~^~  +  4  -         2A         • 

5  —  23-  r  4-2 

39.  11.3-^-— ^  =  2.3 -(5- 7a;) +  --^. 

,'      2x-.3      .4x  +  5       275       ^, 

41.  ;r =  — 9.4. 

Jx  '     X  .25  X 


PROBLEMS 

1.  Separate  the  number  286  into  two  parts  such  that  the 
greater  will  be  2^  times  the  less. 

2.  Separate  the  number  1010  into  three  parts  such  that  the 
second  will  be  -^-  of  the  first  and  the  third  will  be  -^-^-  of  the 
first. 

3.  By  what  number  must  352  be  divided  so  as  to  give  a 
partial  quotient  15  and  the  remainder  7  ? 

4.  What  number  must  be  subtracted  from  both  terms  of 
the  fraction  |-|  to  give  a  fraction  equivalent  to  J-  ? 

5.  Separate  133  into  two  parts  such  that  their  quotient  is  2^. 

6.  Separate  96  into  two  parts  such  that  56  exceeds  two 
thirds  of  the  one  by  as  much  as  the  other  exceeds  16. 

7.  A  boy  is  12  years  old  and  his  sister  8  years  old.  In  how 
many  years  will  the  boy  be  f  as  old  as  his  sister  ? 

8.  Two  thirds  a  man's  age  now  equals  |-  his  age  30  yearg 
ago    What  is  his  age  ? 


64  SECOND  COUKSE  IN  ALGEBRA 

9.  The  square  of  a  certain  number  is  4  greater  than  two 
thirds  of  the  product  of  the  next  two  consecutive  numbers. 
Find  the  number. 

10.  The  length  of  a  certain  rectangle  is  2|-  times  the  width.. 
If  it  were  10  yards  shorter  and  Ij  yards  wider,  its  area  would 
be  1260  square  feet  less.  Find  the  dimensions  of  the  rectangle. 

11.  A  square  court  has  f  the  area  of  a  rectangular  court 
whose  length  is  4  yards  greater  and  whose  width  is  3  yards 
less.    Find  the  dimensions  of  the  square  court. 

12.  A  can  do  a  piece  of  work  in  10  days  and  B  in  12  days. 
How  many  days  will  they  both  require  working  together  ? 

Hint.  Let  x  =  number  of  days  required  by  A  and  B  together.  Then 
-  =  fractional  part  of  the  piece  of  work  that  they  can  do  in  1  day. 

13.  A  can  do  a  piece  of  work  in  10  days  and  B  in  15  days. 
After  they  have  worked  together  5  days,  how  many  days  will 
A  require  to  finish  the  work  ? 

14.  A  tank  has  a  supply  pipe  which  fills  it  in  4  hours  and 
a  waste  pipe  which  empties  it  in  6  hours.  If  the  tank  is  empty 
and  both  pipes  are  opened,  how  much  time  must  elapse  before 
the  tank  is  filled  ? 

15.  A  tank  has  a  supply  pipe  which  fills  it  in  4  hours  and 
two  waste  pipes  which  empty  it  in  6  and  8  hours  respectively. 
If  the  tank  is  full  and  all  three  pipes  are  open,  how  much 
time  will  be  required  to  empty  the  tank  ? 

16.  If  all  three  pipes  of  Problem  15  were  outlet  pipes,  how 
long  would  be  required  to  empty  the  tank  ? 

17.  If  in  Problem  15  the  supply  pipe  had  been  closed  after 
4  hours,  how  much  more  time  would  have  been  required  to 
empty  the  tank? 

18.  The  diameter  of  the  earth  is  3|  times  that  of  the  moon, 
and  the  difference  of  the  two  diameters  is  57 GO  miles.  Find 
each  diameter  in  miles. 


FEACTIONS  65 

19.  The  diameter  of  the  sun  is  3220  miles  greater  than  109 
times  the  diameter  of  the  earth,  and  the  sum  of  the  two  diame- 
ters is  874,420  miles.    Find  each  diameter  in  miles. 

20.  The  diameter  of  Jupiter  is  lOj-J  times  the  diameter 
of  the  earth,  and  the  sum  of  their  diameters  is  94,320  miles. 
Find  each  diameter  in  miles. 

21.  A  man  who  can  row^  miles  per  hour  in  still  water 
rows  up  a  stream  the  rate  of  whose  current  is  Ij-  miles  per 
hour.  After  rowing  back  he  finds  that  the  entire  journey  re- 
quired 10  hours.   Find  the  time  required  for  the  trip  upstream. 

22.  A  man  who  can  row  4 J  miles  per  hour  in  still  water 
finds  that  it  requires  6i  hours  to  row  upstream  a  distance 
which  it  requires  2|-  hours  to  row  down.  Find  the  rate  of  the 
current. 

23.  A  passenger  train  whose  rate  is  40  miles  per  hour  leaves 
a  certain  station  2  hours  and  45  minutes  after  a  freight  train. 
The  passenger  train  overtakes  the  freight  in  5  hours  and  15 
minutes.    Find  the  rate  of  the  freight  train  in  miles  per  hour. 

24.  A  man  invests  a  part  of  $8000  at  5%  and  the  remainder 
at  4%.  If  the  yearly  interest  on  the  whole  investment  is  $345, 
how  much  was  invested  at  each  rate  ? 

25.  A  man  invests  $6800  in  two  parts  :  the  first  part  at  5%, 
and  the  second  part  at  4%.  If  the  average  rate  of  interest  is 
4f  %,  find  the  amount  of  each  investment. 

26.  Two  thousand  dollars  of  Mr.  A's  income  is  not  taxed. 
All  of  his  income  over  that  amount  is  taxed  2%,  and  all  above 
$10,000  is  taxed  2%  in  addition.  He  pays  a  tax  of  $180. 
What  is  his  income? 

27.  How  much  water  must  be  added  to  a  gallon  of  alcohol 
95%  pure  so  as  to  make  a  mixture  10%  pure  ? 

Hint.   Let  w  =  the  number  of  gallons  of  water  to  be  added. 

Then  nkll  =  i?-,  etc. 

1  +  w       100 


66  SECOND  COURSE  IN  ALGEBRA 

28.  How  many  ounces  of  alloy  must  be  added  to  45  ounces 
of  silver  to  make  a  composition  60%  silver  ? 

29.  It  is  desired  to  mix  coffee  which  sells  for  twenty-five 
cents  per  pound  with  coffee  which  sells  for  thirty-five  cents 
per  pound  so  as  to  obtain  a  10-pound  mixture  which  may  be 
sold  at  the  same  profit  for  thirty-two  cents  per  pound.  In  what 
ratio  must  the  parts  be  taken  f  r^m  each  grade  ? 

30.  Milk  of  a  certain  grade  is  known  to  test  20%  cream. 
How  much  water  must  be  added  to  25  gallons  of  this  milk  to 
make  a  mixture  18%  cream  ? 

31.  Grun  metal  of  a  certain  grade  is  composed  of  16%  tin 
and  84%  copper.  How  much  tin  must  be  added  to  410  pounds* 
of  this  gun  metal  to  make  a  composition  18%  tin  ? 

Hints.  Since  the  composition  is  16%  tin,  ^W  •  410  =  the  number  of 
pounds  of  tin  in  the  first  composition. 

Let  X  =  the  number  of  pounds  of  tin  to  be  added. 

.  Then \-  x  =  the  number  of  pounds  of  tin  in  the  sec- 
ond composition, 

and  410  +  X  =  the  number  of  pounds  of  both  metals  in 

the  second  composition. 

16.410 

+  X 

Therefore  ■'     =  — ,  etc. 

410  +  X         100 

32.  How  many  "gallons  of  alcohol  90%  pure  must  be  mixed 
with  12  gallons  of  alcohol  96%  pure  to  make  a  mixture  93% 
pure  ? 

33.  A  certain  lot  of  pig  iron  contains  93%  pure  iron.  How 
much  pure  iron  must  be  melted  with  10  tons  of  pig  iron  to 
make  iron  98%  pure  ? 

34.  The  arms  of  a  lever  are  5  feet  and  6  feet  in  length 
respectively.  Excluding  the  weight  of  the  beam,  what  weight 
on  the  shorter  arm  will  balance  70  pounds  on  the  longer  ? 

Hint.   The  products  of  the  weights  by  their  respective  arms  are  equal. 


FKACTIONS  67 

35.  The  arms  of  a  balanced  lever  are  7  feet  and  12  feet 
respectively,  the  shorter  arm  carrying  a  load  of  36  pounds. 
Find  the  load  on  the  longer  arm. 

36.  If  the  load  on  the  longer  arm  in  Problem  35  be  reduced 
7  pounds,  how  many  feet  from  the  fulcrum  must  a  12-pound 
weight  be  placed  on  the  longer  arm  to  restore  the  balance  ? 

37.  A  beam  12  feet  long  supported  at  each  end  carries  a  load 
of  2  tons  at  a  point  4  feet  from  one  end.  Find  the  load  in  tons 
on  each  support. 

38.  At  what  time  between  3  and  4  o'clock  will  the  hands  of 
a  clock  be  together  ? 

Solution.  The  minute  hand  moves  twelve  times  as  fast  as  the  hour 
hand.    While  the  minute  hand  travels  x  spaces,  the  hour  hand  travels 

—  spaces.    Hence  a;  —  —  equals  the  number  of  spaces  gained  by  the 

minute  hand  in  any  given  time  x. 

In  the  time  from  3  o'clock  until  the  hands  are  together,  the  minute 
hand  must  gain  15  minute  spaces  to  overtake  the  hour  hand. 

X 

Therefore  x  —  —  =  15. 

Whence  x  =  16  y\. 

Hence  the  hands  are  together  16  y\  minutes  after  3  o'clock. 

39.  At  what  time  between  9  and  10  o'clock  are  the  hands  of 
a  clock  together  ? 

40.  At  what  times  between  3  and  4  o'clock  are  the  hands  of 
a  clock  in  a  straight  line  ? 

41.  At  what  time  between  5  and  6  o'clock  is  the  minute  hand 
10  minute  spaces  ahead  of  the  hour  hand  ?  10  minute  spaces 
behind  the  hour  hand  ? 

42.  At  what  times  between  4  and  5  o'clock  are  the  hands  of 
a  clock  at  right  angles  ? 

43.  If  the  average  height  of  n  boys  is  x  inches,  what  is  the 
sum  of  their  heights  in  yards  ? 


68  SECOND  COURSE  IK  ALGEBKA 

44.  If  3  m  4-  5  packages  weigh  j9  pounds,  wliat  is  the  weight 
of  n  of  them  ? 

45.  If  it  takes  a  man  t  hours  to  do  a  piece  of  work,  what 
portion  of  the  work  can  he  do  in  1  hour  ?  What  portion  of  the 
work  would  n  men  do  in  1  hour  ?  What  portion  would  k  men 
do  in  h  hours  ? 

46.  If  it  takes  n  men  h  hours  to  do  a  piece  of  work,  how 
long  will  it  take  x  men  to  do  it  ? 

47.  A  man  buys  bananas  at  d  cents  a  dozen  and  sells  them 
for  h  cents  each.    What  does  he  gain  on  n  dozen  ? 

48.  A  man  bought  m  articles  for  c  cents  per  hundred.  He 
sold  them  all  for  |10.    How  many  dollars  did  he  lose  ? 

49.  Itn  yards  of  ribbon  cost  c  cents,  find  the  cost  of  x  yards. 

50.  If  «/  yards  of  ribbon  cost  x  cents,  how  many  yards  can 
be  bought  for  d  dollars  ? 

51.  A  man  buys  goods  for  b  dollars  and  sells  them  for  s 
dollars.    What  is  his  per  cent  of  gain  ? 

52.  One  man  can  do  a  piece  of  work  in  d  days,  another  can 
do  the  same  work  in  n  days.  How  many  days  will  it  take  both, 
working  together  ? 

53.  If  it  takes  h  hours  to  mow  m  acres,  how  many  days  of 
10  hours  each  will  it  take  to  mow  n  acres  ? 

54.  A  train  goes  y  yards  in  t  seconds.  If  this  equals  m  miles 
per  hour,  write  an  equation  involving  y,  #,  and  m. 

55.  If  71  men  can  do  a  piece  of  work  in  d  days,  how  many 
men  would  it  be  necessary  to  hire  if  the  work  had  to  be  done 
in  t  days  ? 

56.  A  transport  plying  between  two  ports  is  under  fire  for 
/  feet  of  the  way.  If  she  steams  k  knots  per  hour,  for  how 
many  minutes  is  she  under  fire  ? 

Hint.    1  knot  =  6080  feet. 


CHAPTER  V 

LINEAR  SYSTEMS 

44.  Graphical  solution  of  a  linear  system.  The  construc- 
tion of  the  graph  of  a  single  linear  equation  in  two  unknowns 
or  of  a  linear  system  in  two  unknowns  depends  on  several 
assumptions  and  definitions.    It  is  agreed: 

I.  To  have  at  right  angles  to  each  other  two  lines, 
X^OX^  called  the  jr-axis,  and  F'OY,  called  the  y-axis. 

II.  To  have  a  line  of  definite  length  for  a  unit  of  distance. 

Thus  the  number  2  will  correspond  to  a  distance  of  twice  the 
unit,  the  number  4|  to  a  distance  4^  times  the  unit,  etc. 

III.  That  the  distance  (measured  parallel  to  the  a:-axis) 
from  the  ?/-axis  to  any  point  in  the  paper  be  the  j:-distance 
(or  abscissa)  of  the  point,  and  the  distance  (measured 
parallel  to  the  ?/-axis)  from  the  :r-axis  to  the  point  be  the 
y-distance  (or  ordinate)  of  the  point. 

IV.  That  the  a;-distance  of  a  point  to  the  riglit  of  the 
?/-axis  be  represented  by  a  positive  number,  and  the  x- 
distance  of  a  point  to  the  left  by  a  negative  number ;  also 
that  the  ^-distance  of  a  point  above  the  a:-axis  be  repre- 
sented by  a  positive  number,  and  the  y-distance  of  a  point 
helow  the  a:-axis  by  a  negative  number.  Briefly,  distances 
measured  from  the  axis  to  the  right  or  upward  are  positive-,  to 
the  left  or  downward  are  negative. 

V.  That  every  point  in  the  surface  of  the  paper  corre- 
sponds to  a  pair  of  numbers,  one  or  both  of  which  may  be 
positive,  negative,  integral,  or  fractional. 

69 


70  SECOND  COUKSE  IN  ALGEBRA 

• 

VI.  That  of  a  given  pair  of  numbers  the  first  be  the 
measure  of  the  a^-distanee  and  the  second  the  measure  of 
the  ^-distance. 

Thus  the  point  (2,  3)  is  the  point  whose  x-distance  is  2  and  whose 
^-distance  is  3. 

The  point  of  intersection  of  the  axes  is  called  the  origin. 

The  values  of  the  a;-distance  and  the  ^-distance  are  often 
called  the  coordinates  of  the  point. 

The  relation  between  an  equation  and  its  graph  may  be 
stated  as  follows : 

The  equation  of  a  line  is  satisfied  hy  the  values  of  the 
x-distances  and  the  y-distance  of  any  point  on  that  line. 

Any  pointy  the  values  of  whose  x-distance  and  whose  y- 
distance  satisfy  the  equation,  is  on  the  graph  of  the  equation. 

The  graph  of  a  linear  equation  in  two  unknowns  is  a 
straight  line.  Therefore  it  is  necessary  in  constructing 
the  graph  of  such  an  equation  to  locate  only  two  points 
whose  coordinates  satisfy  the  equation  and  then  to  draw 
through  the  two  points  a  straight  line.  It  is  usually  most 
convenient  to  locate  the  two  points  where  the  line  cuts 
the  axes.  If  these  two  points  are  very  close  together, 
however,  the  direction  of  the  line  will  not  be  accurately 
determined.  This  error  can  be  avoided  by  selectmg  two 
points  at  a  greater  distance  apart. 

The  graphical  solution  of  a  linear  system  in  two  unknowns 
consists  in  plotting  the  two  equations  to  the  same  scale 
and  on  the  same  axes  and  obtaining  from  the  graph  the 
values  of  x  and  y  at  the  point  of  intersection  of  the  lines. 

Through  the  graphical  study  of  equations  we  unite  the 
two  subjects  of  geometry  and  algebra,  which  have  hitherto 
seemed  quite  separate,  and  learn  to  interpret  problems  of 
the  one  in  the  language  of  tke  other. 


LINEAR  SYSTEMS 


Tl 


EXAMPLE 

Solve  graphically  the  system 

3^  _  4?/ +  20  =  0, 

Solution.    Substituting  zero  tor  x  in  S  x  —  4:  y  +  20  =  0,  we  obtain 

y  =  b.    Substituting  zero  for  y,  we  obtain  a:=— 6|.    This  may  be 
expressed  in  tabular  form : 


li  x  = 

0 

-6| 

then  y  = 

5 

■    0 

Similarly,  for  the  equation 
2x  +  2^4-6  =  0  we  obtain 
the  following  table : 


If  x  = 

0 

-3 

then  y  = 

-6 

0 

Then,  constructing  the 
graph  of  each  equation  as 
indicated  in  the  adjacent 
figure,  we  obtain  for  the 
coordinates  of  the  point  of 
intersection  of  the  two  lines 
a;  =  —  4  and  y  =  2. 


Y 

y 

5                           ^^ 

V        ^y 

5             -..^4 

V        ^^      * 

\,yr 

\T\ 

X      ^   3                 X 

/'-«  -♦  A''     "    ' 

j^      3r  2 

4    =^ 

IS        -4 

^ 

A; 

^ 

3 

r 

r 

Solve  graphically : 

2.  aj  -  2/  =  6, 

5x  +  4t/  =  —  15. 

3    aj  4-2^4-11  =  0, 
'  y-x  =  l. 


EXERCISES 


iK  +  2y=4, 
6y4-2x  =  20. 

^  +  4  =  10x, 
2  -  a;  =  0. 

2x4-42/ =  20, 
22/-2  =  x. 


72  SECOND  COURSE  IN  ALGEBRA 

x  +  y  =  ^,  x-y  =  b, 

*  2/  + 2  =  0.  3a^-f-2t/  =  6. 

0^  +  5  =  -32/,  3a^-42/  =  33, 

•  62/  +  2£c-ll  =  0.  4«  +  32/=-6. 

«H-?/  =  4,  2x-4ty  =  % 

'  x  +  2y  =  7.  x-2y=S. 

13.  In  each  of  the  first  three  exercises  will  the  values  of  the 
X-  and  ^/-distances  of  the  point  of  intersection  of  the  two  lines, 
as  obtained  from  the  graph,  satisfy  the  equation  obtained  by- 
adding  the  two  given  equations  ? 

14.  Graph  the  equation  x  —  2  y  =  S.  Then  multiply  both 
members  by  3  and  graph  the  resulting  equation.  Compare  the 
two  graphs.  Th^n  use  —  2  as  a  multiplier  and  graph  the  result- 
ing equation.  Compare  the  three  graphs.  What  conclusion 
seems  warranted? 

15.  What  are  the  coordinates  of  the  origin  ? 

16.  Is  a  graphical  solution  of  a  linear  system  ever  impossible? 
Give  an  example. 

17.  What  is  the  form  of  the  equation  of  a  line  parallel  to 
the  £c-axis  ?  the  ^/-axis  ? 

18.  The  boiling  point  of  water  on  a  centigrade  thermometer 
is  marked  100°,  and  on  a  Fahrenheit  212°.  The  freezing  point 
on  the  centigrade  is  zero  and  on  the  Fahrenheit  32°.  Con- 
sequently a  degree  on  one  is  not  equal  to  a  degree  on  the  other, 
nor  does  a  temperature  of  60°  Fahrenheit  mean  60°  centigrade. 
Show  that  the  correct  relation  is  expressed  by  the  equation 
C  =  f  (F  —  32),  where  C  represents  the  number  of  degrees 
centigrade  and  F  the  number  of  degrees  Fahrenheit. 

19.  Construct  a  graph  of  the  equation  in  Exercise  18,  using 
C  and  F  as  abscissa  and  ordinate.  Can  you,  by  means  of  this 
graph,  express  a  centigrade  reading  in  degrees  Fahrenheit 
and  vice  versa? 


LINEAR  SYSTEMS  73 

20.  By  means  of  the  graph  drawn  in  Exercise  19  express 
the  following  centigrade  readings  in  Fahrenheit  readings  and 
vice  versa:    (a)60°C.;    (^)  150°F. ;    (c)-20°C.;    (d)-SO°¥. 

21.  From  the  graph  determine  what  reading  means  the  same 
temperature  on  both  scales. 

45.  Elimination.  In  order  to  find  values  of  x  and  2/ 
which  satisfy  the  equation 

32:+2y=20,  (1) 

when  we  know  that 

^  =  22:  +  3,  (2) 

we  may  substitute  for  7/  in  the  first  equation  the  value  of 
y  from  the  second,  obtainmg  the  single  equation  in  x, 
Sx-h2(^2x-{-S)=20,    or    72^  =  14. 

The  process  by  which  we  have  obtained  one  equation 
containing  one  unknown  from  the  two  equations  (1)  and 
(2)  each  of  which  contains  two  unknowns  illustrates  one 
method  of  elimination. 

In  general,  the  process  of  deriving  from  a  system  of  n 
equations  a  system  of  n  —  1  equations,  containing  one 
variable  less  than  the  original  system,  is  called  elimination. 

For  example,  when  n  =  2,  if  we  have  a  system  of  two  equations 
in  two  unknowns,  the  process  of  elimination  leads  to  one  equation 
in  one  unknown.  Since  we  can  always  solve  such  an  equation,  it 
appears  that  we  can  solve  a  system  of  two  equations  in  two  unknowns 
whenever  it  is  possible  to  eliminate  one  of  the  unknowns.  We  shall 
see  that  only  in  certain  exceptional  cases  is  elimination  impossible. 
This  is  either  because  more  than  one  unknown  is  removed  when 
we  try  it  or  because  the  result  of  attempted  elimination  is  not  an 
equation. 

Only  two  methods  of  solution  will  be  considered  —  that 
involving  elimination  by  substitution  and  that  involving 
elimination   by  addition   or  subtraction. 


74  SECOND  COUKSE  IN  ALGEBEA 

46.  Solution  by  substitution.  The  method  of  solving  a 
system  of  two  linear  equations  by  substitution  is  illustrated 
in  the 

EXAMPLE 

Solve  the  system/^  ~^^,^^^^;,  W 

^         18  0-  + 11 2/ =  18.  (2) 

Solution.    From  (1),  a;  =  13  ?/  +  81.  (3) 

Substituting  this  value  for  x  in  (2), 

8  (13  2^ +  31) +  11^  =  18. 
Simplifying,  104  z/  +  248  +  11  ?/  =  18. 
Combining  terms,  115  y  =  —  230. 

Whence  y  =—2. 

Substituting  —  2  tor  y  in  (3),  and  solving, 

x  =  6. 
Check.    Substituting  5  for  x  and  —  2  for  y  in  (1)  and  (2), 
5-13(-2)  =  31,   or  31  =  81, 
and  8-5 +11  (-2)  =  18,   or   18=18. 

The  method  of  the  preceding  solution  is  stated  in  the 
following 

Rule,  Solve  either  equation  for  the  value  of  one  unJmown  in 
terms  of  the  other. 

Substitute  this  value  for  the  unknown  which  it  represents^  in 
the  equation  from  which  it  was  not  obtained^  and  solve  the 
resulting  equation. 

In  the  simplest  of  the  preceding  equations  which  contains 
both  unknowns,  substitute  the  definite  value  just  founds  and 
solve,  thus  obtaining  a  definite  value  for  the  other  unknown. 

Check.  Substitute  for  each  unknown  in  both  original  equa- 
tions its  value  as  found.  If  the  resulting  equations  are  not 
obvious  identities,  simplify  them  until  they  become  so. 


3x 

-22/: 

=  18, 

42/ 

+  Sx. 

=  0. 

8m 

-Sn 

+  6/ 

2 

4  m 

-1  = 

-Sn. 

LINEAR  SYSTEMS  75 

EXERCISES 

Solve  by  substitution : 

Sx-Sy  =  20,  5. 

X  —  61/  =  0. 

2x  +  52/  =  8, 

^    2(x  +  y)+Stj  =  4:,  4.t-2n  =  lS, 

'  5==x-hy-  ^-  202^  =  7/1  +  63. 

^    16x  +  7  =  152/,  6a;  +  38  =  12y, 

•  4a:4-52/  =  0.  ^'  4:X-Sy  =  0. 

47.  Solution  by  addition  or  subtraction.  The  method  of 
solving  a  system  of  two  linear  equations  by  addition  or 
subtraction  is  illustrated  in  the 

EXAMPLE 

c  1      .1.  .       fl3x  +  3y  =  U,  (1) 

Solve  the  system |^^_^^^^^^;  (2) 

Solution.   First  eliminate  y,  as  follows : 

(1)  •  2,  26  a;  +  6  ?/  =  28  (3) 

(2).  3,  21a:-6.y:^66  (4) 

(3) +  (4),  47  a:  =94  (5) 

(5)  ^47,  a;  =  2.  (6) 

Substituting  2  for  x  in  (2),  we  obtain  y  =  —  4. 
Check.    Substituting  2  for  x  and  —  4  for  y  in  (1)  and  (2), 
26  - 12  =  14,   or   14  =  14, 
and  14  +  8  =  22,   or   22  =  22. 

Before  starting  to  eliminate,  the  system  should  always 
be  inspected  carefully  to  determine  which  unknown  can 
be  removed  most  conveniently.  In  this  case  the  elimina- 
tion of  y  is  the  simpler,  because  it  involves  multiplication 
•by  smaller  numbers  than  does  the  elimination  of  x. 


76  SECOND  COUESE  IN  ALGEBRA 

The  method  of  the  preceding  solution  is  stated  in  the 
following 

Rule.  If  necessary^  multiply  each  member  of  the  first  equa- 
tion hy  a  number^  and  each  member  of  the  second  equation  by 
another  number,  such  that  the  coefficients  of  the  same  unhnoum 
in  the  resulting  equations  will  be  numerically  equal. 

If  these  coefficients  have  like  signs,  subtract  one  equation 
from  the  other  ;  if  they  have  unlike  signs,  add ;  then  solve  the 
equation  thus  obtained. 

In  the  simplest  of  the  preceding  equations  which  contairis 
both  unknowns,  substitute  the  value  just  found  and  solve  for 
the  other  u7iknown. 

Check.    As  on  page  74. 

ORAL  EXERCISES 

Solve  the  following  systems  : 

'  x-y  =  2,  '  2x  +  y  =  ^.  *2ic-3?/  =  9. 

g    a^  +  27/  =  3,       ^    2aj+32/  =  8,         ^hx-6y  =  7, 
'  X  -\-  y  =  2.  '  x  —  y  =  1.  '  Ax  —  Sy  =  2. 

48.  Special  cases.  The  equation  x  +  y  =  10  has  as  roots 
any  set  of  two  numbers  whose  sum  is  10.  li  x-\-  y  =  5  is 
taken  as  the  other  equation  of  a  system,  one  can  see  im- 
mediately that  tlie  two  equations  have  no  set  of  roots  in 
common,  since  the  sum  of  two  numbers  cannot  be  10  and 
5  at  the  same  time. 

A  system  of  equations  which  has  a  common  set  of 
roots  is   called   a   simultaneous  system. 

A  system  of  equations  which  does  not  have  a  common 
set  of  roots  is  called  inconsistent  or  incompatible. 


LINEAE  SYSTEMS  7T 

The  attempt  to  solve  an  incompatible  system  results  in 
getting  rid  not  only  of  one  but  of  both  unknowns  and 
leads  to  a  statement  in  the  form  of  an  equation  which  is 
false. 

Consider  x  -\-  y  =  10,  (1) 

x  +  y  =  5.  (2) 

(1)  -  (2),  0  =  5,  which  is  false. 

If,  on  the  other  hand,  the  equation  x-\-i/  =  10  is  taken 
for  one  equation  of  a  system  and  2a:  +  2?/=20  for  the 
other,  it  appears  that  any  set  of  numbers  which  satisfies 
one  equation  satisfies  also  the  other,  since  if  the  sum  of 
two  numbers  is  10,  the  sum  of  twice  those  numbers  is  20, 
and  any  one  of  the  countless  sets  of  roots  of  one  equa- 
tion is  a  set  of  roots  of  the  other.  In  fact,  the  second 
equation  may  be  obtained  from  the  first  by  multiplying 
each  member  by  2. 

If  one  equation  of  a  system  can  be  obtained  from  one 
or  more  of  the  other  equations  of  the  system  by  applica- 
tion of  one  or  more  of  the  axioms,  it  is  called  a  derived  or 
dependent  equation.  If  it  cannot  be  so  obtained,  it  is  called 
independent. 

Thus  equations  (1)  and  (2)  in  the  example  on  page  75  are 
independent,  while  equation  (5)  is  derived  from  them. 

An  attempt  to  eliminate  one  unknown  from  a  system 
of  two  equations  in  two  unknowns  which  are  not  inde- 
pendent results  in  gettmg  rid  not  only  of  both  unknowns 
but  of  the  constant  terms  as  well,  so  that  only  the  identity 
0  =  0  remains. 

Thus  x  +  y  =  10,  (1) 

2x  +  2y  =  20.  (2) 

(1).2,  2a; +  2^  =  20.  (8) 

(2) -(3),  0  =  0. 


78  SECOND  COURSE  IN  ALGEBRA 

ORAL  EXERCISES 

Which  of  the  following  systems  are  incompatible,  which  are 
simultaneous,  and  which  are  dependent  ? 

x  +  y  =  2,  ^x-22j=4:, 

'x-\-y  =  l.  *5ic  — 102/  =  4. 

3x  +  32/  =  6,  ?>x-y  =  2, 

^'  x  +  y  =  2.  •  9^^-32/ =  6. 

2  a; +  2/ =  4,  3^ -2/ =  2, 

"*•  4a^  +  23/  =  l.  *  9a:  +  32/  =  6. 

0^  +  2/  =  3,  2t/-^  +  6  =  0, 

*•  x-2/  =  l.  •  3a; -67/ -21  =  0. 

2a;4-32/  =  4,  2a3  =  72/-5, 

4a; +  62/ =  8.  *  8a;  =  282/ 4- 3. 

EXERCISES 

Solve  the  following  systems  : 


2a;   ,               _ 

11          1 

3    +^=      ^' 

6.  "^      ^          6 
2      3_4 

5a; -3^/  =  105. 

3r       7        s 

a;      2/~3* 

4        2      12' 

1         1 

r+8=-2s. 

Hint  :  Solve  first  for  -  and  - 

X         y 

32/ +  1      ^  +  22 

=  3 

4        12 

> 

2r-h4s      38 

«-2y  =  l. 

2r-s   ~  3  ' 
^      ^  -  0 

.5x  +  .73,  =  |, 

7*          S 

.8a;-.22/  =  3f. 

w  -h  2  71      2m  — 

71 

5 

2m  +  Sn-2      4 

__  ;:::: 

— , 

5                 10 

2 

m+^+6         3 

w,  +  71      m  —  n 

71 

4               7 

'2* 

LINEAR  SYSTEMS  79 


x-^      y        '  3              -11 

'  X  —  ^y  _2  5                 5 

5      '~n'  2x  +  y  =  l. 

2.^h-\-l-.lk  ^  X      y      6j/      2x 

k-l(^  +  h  '                   5  "^2        2   "^  5 


.8A-2.2"^35-5.5A:        '      ^^-       4  2 


2h-Sk 


=  3, 


X 


1 


4-2y      4 


n.    ^-^^  5^±^4.2=   ^^ 


^—  =  39.  ^^-.  +  7      17  +  . _Q 

3  «  5-« 


Solve  for  ic  and  y : 

jg    5a; +  4?/  =  10a +  4, 
jc  —  2  ay  =  0. 

7x  +  5y  =  21c, 

2  acc  — 3^*^  =  7, 
*   5aic  +  7Z>y  =  3. 

Zx      y 


2-2  =  6, 

19.  "       "     ■ 

^  +  1_  =  13 

2a      3c  ■  Sax-6by  =  5c. 


20. 

a      Sa 

X       y 

Za      a      1 

^      y~2' 

21. 

4a                       / 
a 

22. 

X       y      a  -\-h 

ah          ab 

a^-Jy" 

X  —  y  = ; 

^          ah 

oo 

2  ace  ~  4  6?/  =  7  c, 

^0  SECOND  COURSE  IN  ALGEBRA 

^^    ax  +  hy  =  c,  ^^    ax  +  hy  =  c, 

kax  +  khy  —  ck.  '  dx  -\-fy  =  ff. 

26.  Show  that  if  af—hd  =  0,  the  equations  in  Exercise  25 
are  inconsistent,  unless  they  form  a  dependent  system. 

27.  Solve  the  system  of  Exercise  17  for  a  and  h  in  terms 
of  X  and  y. 

28.  Solve  the  system  of  Exercise  23  for  a  and  b  in  terms 
of  Xy  y,  and  c. 

49.  Equations  in  several  unknowns.  We  have  already 
seen  that  the  equation  x-{-y  =  10  is  satisfied  by  an  un- 
limited number  of  sets  of  roots,  since  there  is  an  infinity 
of  pairs  of  numbers  whose  sum  is  10. 

An  equation  or  a  system  of  equations  which  is  satisfied  by 
an  infinite  number  of  sets  of  roots  is  said  to  be  indeterminate. 

If  a  simultaneous  system  is  satisfied  by  only  a  limited 
number  of  set.*  of  roots,  it  is  said  to  be  determinate. 

The  system  x  +  ?/  =  10,  a;  —  y  =  2,  is  determinate  and  has  the  set  of 
roots  (6,  4).    The  system  2  x  +  2  y  =  20,  x  +  y  =  10,  is  indeterminate. 

When  we  consider  systems  of  equations  in  three  un- 
knowns, the  question  arises  whether  two  such  equations 
form  a  determinate  system.    For  example,  the  equation 

x  +  y-^z=\0  (1) 

is  satisfied  by  an  infinite  number  of  sets  of  roots.  If  we 
consider  a  system  consisting  of  (1)  and 

^-(y  +  ^)=2,  (2) 

it  appears  from  inspection  that  the  system  is  satisfied  if 
x  =  Q  and  ?/  -f  2  =  4.  But  the  equation  y  -\-z  =  ^  is  satis- 
fied by  an  infinite  number  of  sets  of  roots.  Hence  equar 
tions  (1)  and  (2)  form  an  indeterminate  system. 


LINEAR  SYSTEMS  81 

If,  however,  we  adjoin  a  third  equation  to  the  system,  as 
y-z=%  (3) 

it  becomes  determinate,  since  ?/  +  ^  =  4  and  y  —  z=2  are 
satisfied  only  by  the  set  of  numbers  3,  1.  It  is  usually 
true  that  three  equations  in  three  unknowns  form  a  deter- 
minate system. 

In  general,  when  the  number  of  unknowns  in  a  system 
of  linear  equations  exceeds  the  number  of  equations,  the 
system  is  indeterminate.  If  the  number  of  equations 
equals  the  number  of  unknowns,  the  system  is  usually 
determinate  and  simultaneous.  If  the  number  of  equations 
exceeds  the  number  of  unknowns,  the  system  is  usually 
inconsistent.  There  are  many  special  cases  wliich  arise  in 
the  study  of  hnear  systems  in  n  unknowns,  corresponding 
to  those  mentioned  for  two  unknowns  in  section  48,  but 
they  become  very  complicated  for  larger  values  of  n^  and. 
a  thorough  study  of  them  is  quite  beyond  the  scope  of 
this  text. 

Note.  It  is  not  a  little  remarkable  that  the  writings  of  the  first 
great  algebraist,  Diophantos  of  Alexandria  (about  a.d.  275),  are  de- 
voted almost  entirely  to  the  solution  of  indeterminate  equations ; 
that  is,  to  finding  the  sets  of  related  values  which  satisfy  an  equa- 
tion in  two  unknowns,  or,  perhaps,  two  equations  in  three  unknowns. 
We  know  practically  nothing  of  Diophantos  himself,  except  the 
information  contained  in  his  epitaph,  which  reads  as  follows : 
".Diophantos  passed  one  sixth  of  his  life  in  childhood,  one  twelfth 
in  youth,  one  seventh  more  as  a  bachelor;  five  years  after  his  mar- 
riage a  son  was  born  who  died  four  years  before  his  father,  at  half 
his  father's  age."  From  this  statement  the  reader  was  supposed 
to  be  able  to  find  at  what  age  Diophantos  died.  As  a  mathemati- 
cian Diophantos  stood  a;lone,  without  any  prominent  forerunner 
or  disciple,  so  far  as  we  know.  His  solutions  of  the  indetermi- 
nate equations  were  exceedingly  skillful,  but  his  methods  were  so 
obscure  that  his  work  had  comparatively  little  influence  upon  later 
mathematicians. 


82       SECOND  COURSE  IN  ALGEBRA 

50.  Determinate  systems.  The  method  of  obtaining  the 
set  of  roots  of  a  determinate  system  in  three  unknowns  is 
illustrated  in  the 

EXAMPLE 

(x-\-62j-5z  =  21,  (1) 

Solve  the  system  iSx  —  Sy-\-z=—  5,  (2) 

[5x-7y-^2z  =  4:.  (3) 

Solution.   First  eUminate  one  unknown,  say  z,  between  (1)  and  (2) : 

X  +  6  3/  -  5  2  =  21.  (1) 

(2). 5,  15x-^0y  +  5z=-2o.  (4) 

(l)  +  (4),  lQx-Uij=-4:.  (5) 

Now  eUminate  z  between  (2)  and  (3) : 

(2) -2,  Qx-lQy  +  2z=-10.  (6) 

5x-7y  -\-2z  =  ^.  (3) 

(6)-(3),  x-9y=-U.  (7) 

The  equations  (5)  and  (7)  contain  the  same  two  unknowns  x  and  y. 
lQx-Uy=-^.  (5) 

(7) -16,  16  a: -144?/ =-224.  (8) 

(5) -(8),  110  y  =  220. 

y  =  2. 

Substituting  in  (7),  a;  =  4. 

Substituting  both  these  values  in  (1), 

4  +  12  -  5  ^  =  21. 
Whence  z=  —  l. 

Check.  Substituting  4  for  x,  2  for  y,  and  —  1  for  2  in  (1),  (2), 
and  (3)  respectively, 

4  +  6-2-5(-l)  =  21,     or     21  =  21. 
3.4-8.2  +  (-l)=-5,   or    -5=-5. 
5.4-7.2  +  2(-l)  =  4,       or       4  =  4. 


LINEAR  SYSTEMS  83 

For  the  solution  of  a  simultaneous  system  of  equations 
in  three  unknowns  we  have  the 

Rule.  From  an  inspection  of  the  coefficients  decide  which 
unknown  is  most  easily  eliminated. 

Using  any  two  equations^  eliminate  that  unknown. 

With  one  of  the  equations  just  u^ed  and  the  third  equa- 
tion again  eliminate  the  same  unknown. 

The  last  two  operations  give  two  equations  in  the  same  two 
unknowns.    Solve  these  equations. 

Substitute  in  the  simplest  of  the  original  equations  the  two 
values  found,  and  solve  for  the  third  unknown. 

Check.  Substitute  the  values  found  in  the  original  equations 
and  simplify  results. 

.    A  system  of  four  independent  equations  in  four  un- 
knowns may  be  solved  as  follows  : 

Use  the  first  and  second  equation,  then  the  first  and 
third,  and  lastly  the  first  and  fourth,  and  eliminate  the 
same  unknown  each  time.  This  gives  a  system  of  three 
equations  in  the  same  three  unknowns,  which  can  be  solved 
by  the  rule  given  above. 

EXERCISES 

Solve  for  x,  y,  and  z  and  check  the  results : 

x  +  ^y  —  ^z  =  2,  x  +  y  +  z  =  l, 

I.  2x  —  y  —  z  =  1,  ^.  X  -{-  y  —  z  =  2, 

Sx-{-  5y-7z=-10.  x-?/  +  ^  =  3. 

2x+Sy-i-4.-z=-14:,  .  x  +  2y -\- z  =  1, 

2.x  —  y  +  3z  =  0,  5.2x-^y  —  z  =  0, 

5x-\-2y  +  z  =  14:.  x-\-2y  -{-z  =  0. 

x+'2y-\-z  =  -l,  2x-y-^5z=0, 

3.  2x-y-}-z=-20,  6.  Sx -^7  y -\- z  =  SS, 

—  X  —  y  —  5  z  =  IS.  X  —  6y  —  z  =  7. 


84  SECOND  COUESE  IN  ALGEBEA  ' 

X      y       z 

X        1^        y 
x-\-y      £_2. 

7/  4-  ^      X ^ 

""4  2  "12' 

Sx  +  2y=12-3z. 


x-6y-\-Sz  =  ^ 

4x-32/  =  «, 

8. 

s  =  £c  -4-  2/, 

2x  =  32/  +  l. 

2x  +  2/  =  S  +  «, 

9. 

05-2^  =  6, 

32/  +  2«  =  aj. 

^_2y  =  10, 

10. 

3^4-4^=-!, 

5  X  —  ^  =  18. 

^  =  3^  +  2, 

11. 

y  =  .x-7i, 

^  =  6i/-l. 

4x-22/  =  0, 

12 

.  6^-82/=-2, 

x^z  =  \\. 

18. 


A«  J[-hy  —  lz  =  2  hk 

19.  A^i/  —  /^ic  +  ^^  =  2  A;^, 

hx  —  hy  +  1^  =  2  hi. 


20. 


2a; +  2/  +  ^  +  ^  =  ^» 
jC  _  1/  _  ;2  +  2  ?^^  =  4, 

X  +  2  7/  — ^  —  'i^  =  0, 

oj  —'2/  +  2  ^  —  i^;  =  1. 


i. 


^3--  +  ^  =  t"'  ^^•4x-22/  +  ^  — =-9 

^  +  "  =  ^^'  2x-32/-2.  +  ^^  = 
35c4.2i/  =  6a-2^, 

14.  x-5«  +  62/  =  2a-ll^  x  +  2/-«  =  ^j 
6a; -82/ =  12  a +  8^.  ^  _|- 2/ -  w;  =  2, 

ic-?/  +  «  +  w;  =  S) 

ax  +  hy  =  0,  2sc-3y-«  +  i^  =  7. 

15.  cx-6«  =  2  6c, 

bx  -\-  az  —  cy  =  b\  ^  ^y  -\-  z  =  l, 

.3x  +  .22/  +  .4.  =  1.9,  23.  ^"^"""^"i' 

16.  .02x  =  .l-.01//-.02^,  x-z-w:^-5, 


a;  +  2/  +  «  =  6. 


z 
—  «  +  !(;  =  0. 


LINEAR  SYSTEMS  .  85 

PROBLEMS 

Express  the  conditions  of  the  following  problems  by  means 
of  simultaneous  systems,  solve,  and  check  the  results : 

1.  The  sum  of  two  numbers  is  109  and  their  difference 
is  49.    Find  the  numbers. 

2.  A  workman  is  hired  for  30  days.  He  is  paid  $3.50  per 
day  and  board,  but  is  credited  with  80  cents  for  each  day  that 
he  does  not  receive  board.  At  the  end  of  the  30  days  he  re- 
ceives $113.    How  many  days  did  he  receive  board  ? 

3.  Thirty -nine  tons  of  material  are  to  be  moved  by  motor 
trucks  and  drays.  It  is  found  that  the  work  can  be  done  in  a 
given  time  either  by  10  trucks  and  6  drays,  or  by  8  trucks  and 
10  drays.   What  is  the  capacity  of  a  truck  and  of  a  dray  ? 

4.  Two  men  travel  from  New  York  to  the  same  station  by 
rail.  It  costs  one  of  them  three  times  as  much  for  excess 
baggage  as  it  costs  the  other.  One  pays  $7.40  in  all,  the  other 
$10.60.    How  much  does  each  pay  for  his  ticket? 

5.  A  man  and  a  boy  can  do  in  18  days  a  piece  of  work 
which  5  men  and  9  boys  can  do  in  3  days.  In  how  many  days 
can  1  man  do  the  work  ?    1  boy  ? 

6.  A  and  B  together  can  do  a  piece  of  work  in  5  days.  If 
they  work  together  3  days  and  A  can  then  finish  the  job  alone 
in  4  days  more,  how  many  days  does  each  require  alone  ? 

7.  Two  sums  are  put  at  interest  at  5%  and  6%  respectively. 
The  annual  income  from  both  together  is  $100.  If  the  first 
sum  had  yielded  1%  more  and  the  second  1%  less,  the  annual 
income  would  have  decreased  by  $2.    Find  each  sum. 

8.  If  ax  -{-by  =  2  is  satisfied  by  x  =  2  and  y  =  S  and  also 
hj  X  =  6  and  y  =  5,  what  values  must  a  and  b  have  ? 

9.  li  2x-\-b't/  =  c  is  satisfied  when  x  =  l  and  y  =  —  l  and 
also  when  x  =  5  and  y  =  4=,  what  values  must  b  and  c  have  ? 


86  SECOND  COUESE  IN  ALGEBRA 

10.  A  sum  of  $4000  is  invested,  a  part  in  5%  bonds  at  90, 
and  the  remainder  in  6%  bonds  at  110.  If  the  total  annual 
income  is  |220,  find  the  sum  invested  at  each  rate. 

11.  The  purity  of  gold  is  measured  in  carats,  18  carats 
meaning  that  18  parts  out  of  24  are  pure  gold.  A  goldsmith 
has  20  ounces  of  pure  gold  which  he  wishes  to  use  in  making 
16-carat  and  10-carat  alloys.  How  much  pure  gold  can  he  use 
for  each  alloy  if  he  makes  39  ounces  in  all  ? 

12.  A  bag  weighing  18  ounces  contains  two  sizes  of  steel 
balls — ounce  balls  and  |^-ounce  balls.  There  are  23  balls  in  all. 
Find  the  number  of  balls  of  each  size. 

13.  If  the  length  of  a  rectangle  is  decreased  by  7  feet  and 
the  breadth  is  increased  by  8  feet,  the  area  is  unchanged.  If 
the  length  is  increased  by  14  feet  and  the  breadth  is  decreased 
by  4  feet,  the  area  is  also  unchanged.  Find  the  dimensions  of 
the  rectangle. 

14.  A  man  has  |4.50  in  dimes  and  quarters.  If  he  has  36 
coins  in  all,  how  many  has  he  of  each  ? 

15.  A  man  has  $6.00  in  quarters,  dimes,  and  nickels.  He 
has  as  many  quarters  as  he  has  dimes,  and  three  times  as  many 
nickels  as  dimes.   How  many  of  each  has  he  ? 

16.  A  is  half  as  old  as  B.  Seven  years  ago  A  was  one  third 
as  old  as  B.   How  old  is  each  now  ? 

17.  A  man  can  walk  4  miles  per  hour.  He  reaches  a  point 
20  miles  from  his  starting  point  in  three  hours,  having  been 
taken  part  of  the  way  by  a  stage  traveling  12  miles  per  hour. 
How  far  did  the  stage  carry  him  ? 

18.  A  man  and  his  two  sons  can  do  a  piece  of  work  in 
t  of  a  day.  The  two  boys  together  can  do  it  in  1|-  days  and 
one  of  them  can  do  it  in  1  day  less  than  the  other.  What 
portion  of  the  work  does  each  do  when  they  work  together? 


LINEAR  SYSTEMS  87 

19.  Two  automobiles  25  miles  apart  travel  toward  each  other 
and  meet  in  1  hour.  If  they  had  both  traveled  in  the  same 
direction,  the  faster  would  have  overtaken  the  slower  in  5  hours. 
Find  the  rate  of  each. 

20.  An  aeroplane  travels  a  certain  distance  in  3  hours.  If 
the  distance  had  been  half  again  as  great,  the  aeroplane  would 
have  been  forced  to  travel  50  miles  per  hour  faster  in  order  to 
cover  it  in  the  same  time.  Find  the  distance  and  the  speed  of 
the  aeroplane. 

21.  One  angle  of  a  triangle  is  twice  another,  and  their  sum 
equals  the  third.  Find  the  number  of  degrees  in  each  angle  of 
the  triangle.  ■ 

22.  The  sum  of  three  numbers  is  108.  The  sum  of  one  third 
the  first,  one  fourth  the  second,  and  one  sixth  the  third  is  25. 
Three  times  the  first  added  to  four  times  the  second  and  six 
times  the  third  is  504.    Find  the  numbers. 

.  23.  The  sum  of  three  numbers  is  217.  The  quotient  of  the 
first  by  the  second  is  5,  which  is  also  the  quotient  of  the  second 
by  the  third.   Find  the  numbers. 

24.  If  the  tens'  and  units'  digits  of  a  three-digit  number  be 
interchanged,  the  resulting  number  is  27  less  than  the  given 
number.  If  the  same  interchange  is  made  with  the  tens'  and 
hundreds'  digits,  the  resulting  number  is  180  less  than  the 
given  number.   The  sum  of  the  digits  is  14.    Find  the  number. 

25.  In  1  hour  a  tank  which  has  three  intake  pipes  is  filled 
seven-eighths  full  by  all  three  together.  The  tank  is  filled  in 
1^  hours  if  the  first  and  second  pipes  are  open,  and  in  2  hours 
and  40  minutes  if  the  second  and  third  pipes  are  open.  Find 
the  time  in  hours  required  by  each  pipe  to  fill  the  tank. 

26.  The  sum  of  two  sides  which  meet  at  one  of  the  vertices 
of  a  quadrilateral  is  20  feet.  The  sum  of  the  two  which  meet 
at  the  next  vertex  is  27  feet.    The  sums  of  the  two  pairs  of 


88  SECOND  COUESE  IN  ALGEBRA 

opposite,  sides   are  23  feet  and  29  feet  respectively.    Find 
each  side.    (Two  solutions.) 

27.  Two  chairs  cost  h  dollars.  The  first  cost  m  cents  more 
than  the  second.    Find  the  cost  of  each  in  cents. 

28.  Find  two  numbers  whose  sum  is  a  and  whose  difference 
is  h. 

29.  Find  two  numbers  whose  sum  \s>  a-\-h  and  whose  differ- 
ence \s,  a  —  h.  ' 

30.  Two  relays  of  messengers  carry  a  message  k  miles.  The 
first  relay  travels  c  miles  further  than  the  second.  How  far 
does  each  go? 

31.  A  man  has  a  dollars  and  b  cents  in  dimes  and  quarters. 
If  he  has  c  coins  in  all,  how  many  of  each  kind  has  he  ?  ^ 

32.  A  man  has  a  dollars  in  quarters  and  nickels,  with  b 
more  quarters  than  nickels.    How  many  of  each  has  he  ? 

33.  A  and  B  together  can  do  a  piece  of  work  in  m  days. 
B  works  G  times  as  fast  as  A.  How  many  days  does  each 
require  to  do  the  work  alone  ? 

34.  A  man  rows  m  miles  downstream  in  t  hours  and  returns 
in  a  houi's.  Find  his  rate  in  still  water  and  the  rate  of  the 
river. 

35.  Two  contestants  run  over  a  440-yard  course.  The  first 
wins  by  4  seconds  when  given  a  start  of  200  feet.  They  finish 
together  when  the  first  is  given  a  handicap  of  40  yards.  Find 
the  rate  of  each  in  feet  per  second. 

36.  A  train  leaves  M  two  hours  late  and  runs  from  M  to  P  at 
60  %  more  than  its  usual  rate,  arriving  on  time.  If  it  had  run 
from  M  to  P  at  25  miles  per  hour,  it  would  have  been  48  min- 
utes late.    Find  the  usual  rate  and  the  distance  from  M  to  P. 

37.  A  train  leaves  M  30  minutes  late.  It  then  runs  to  N  at 
a  rate  20%  greater  than  its  usual  rate,  arriving  6  minutes  late. 
Had  it  run  15  miles  of  the  distance  from  ilf  to  iV  at  the  usual 


LINEAE  SYSTEMS  89 

rate  and  the  rest  of  the  trip  at  the  increased  rate,  it  would 
have  been  12  minutes  late.  Find  the  distance  from  M  to  N 
and  the  usual  rate  of  the  train. 

38.  It  is  desired  to  have  a  10-gallon  mixture  of  45%  alcohol. 
Two  mixtures,  one  of  95%  alcohol  and  another  of  15%  alcohol, 
are  to  be  used.  How  many  gallons  of  each  will  be  required  to 
make  the  desired  mixture  ? 

Hint.   Let  x  and  y  =  the  number  of  gallons  of  95%  and  15%  alcohol 

respectively.   Then  —. '- — -  =  — ,  and  x  4-  y  =  10. 

^  10  100  ^ 

39.  A  chemist  has  the  same  acid  in  two  strengths.  Eight 
liters  of  one  mixed  with  12  liters  of  the  other  gives  a  mixture 
84%  pure,  and  3  liters  of  the  first  mixed  with  2  liters  of  the 
second  gives  a  mixture  86%  pure.  Find  the  per  cent  of  purity 
of  each  acid. 

40.  When  weighed  in  water  the  crown  of  Hiero  of  Syracuse, 
which  was  part  gold  and  part  silver,  and  which  weighed  20 
pounds  in  air,  lost  1^  pounds.  How  much  gold  and  how  much 
silver  did  it  contain? 

Hint.  When  weighed  in  water  19^  pounds  of  gold  and  10^  pounds 
of  silver  each  lose  1  pound. 

41.  Find  the  positive  integers  which  satisfy  the  equation 
5x  +  22/  =  42. 

4.2  —  9  y 

Solution.  X  = -• 

5 

42  —  2  y 

If  X  is  to  be  a  positive  integer,  -^ — -  must  be  integral ;   that 

o 
is,  42  —  2  y  must  be  a  positive  integral  multiple  of  5.    Hence  y  can 
only  have  the  values  1,  6,  11,  and  16.    The  corresponding  values  of 
X  are  8,  6,  4,  and  2. 

The  various  related  sets  of  integTal  values  of  the  unknowns  which 
satisfy  an  equation  may  be  effectively  represented  to  the  eye  by  the 
graph  of  the  equation.  Since  the  equation  5  re  +  2  y  =  42  has  the 
integral  sets  of  roots  (8,  1),  (6,  6),  etc.,  the  line  of  which  this  is 
the  equation  passes  through  the  points  whose  coordinates  are  these 

RE 


90  SECOND  COURSE  IN  ALGEBEA 

sets  of  integers.  If  the  line  does  not  enter  the  first  quadrant,  we 
can  see  at  a  glance  that  the  corresponding  equation  has  no  positive 
integral  sets  of  roots. 

42.  Solve  in  positive  integers  7x  -|-  2y  =  36,  and  illustrate 
the  result  graphically. 

43.  In  hov^  many  ways  can  a  debt  of  ^73  be  paid  with  5-dollar 
and  2-dollar  bills  ?    Illustrate  the  result  graphically. 

44.  A  man  buys  calves  at  |6  each  and  pigs  at  |4  each, 
spending  |72.    How  many  of  each  did  he*  buy? 

45.  In  liow  many  ways  can  |1.75  be  paid  in  quarters  and 
nickels  ? 

46.  A  farmer  sells  some  calves  at  |6  each,  pigs  at  $3  each, 
and  lambs  at  |4  each,  receiving  for  all  |126.  In  how  many 
ways,  could  he  haye  sold  32  animals  at  these  prices  for  the 
same  sum  ?  Determine  the  number  of  animals  in  the  various 
groups. 

Hint.  Eorm  two  equations  in  three  unknowns.  Then  eliminate  one 
of  the  unknowns. 

47.  In  how  many  ways  can  a  sum  of  $2.40  be  made  up  with 
nickels,  dimes,  and  quarters,  on  the  condition  that  the  number 
of  nickels  used  shall  equal  the  number  of  quarters  and  dimes 
together  ?    Determine  the  various  groups. 


CHAPTER  VI- 

EXPONENTS 

51.  Fundamental  laws  of  exponents.  The  four  laws  of 
exponents  used  in  the  preceding  chapters  are: 

I.  Law  of  Multiplication, 

If  a  and  b  are  positive  integers,  we  have 
x'^  =  X  '  X  '  X  '  X  '  '  •  to  a  factors, 
and  x^  =  X- '  X  '  X  •  '  '  to  h  factors. 

Hence    x^  *  x^  =(x  •  x  *  x  -  -  -to  a  factors)  x  (^x  -  x  >  x  -  - » 
to  b  factors) 
=  X'X'X'-' to  a-^b  factors 
_  ^a+6  \yj  ^i^Q  definition  of  an  exponent. 

II.  Law  of  Division, 

Xa-^  X^  =  X°-^. 

Again,  if  a  and  b  are  positive  integers,  we  have 
.      x^     X'  X  '  X  •  '  '  to  a  factors 

X^^X^=—:  —  = —T 

x'^     X'  X  '  X  '  •  '  to  0  factors 

If  b  is  less  than  a,  the  b  factors  of*  the  denominator  may 
be  canceled  with  b  factors  of  the  numerator,  leaving  a  —  b 
factors  in  the  numerator. 

x^ 
Hence  —  =  x^~^. 

x^ 

•  91 


92  SECOND  COUESE  IN  ALGEBRA 

If  h  is  greater  than  a, 


III.  Law  of  Involution,  or  raising  to  a  power, 

As  before,  if  a  and  h  are  positive  integers,  we  have 
(x^y  =  x^  '  x^  '  x^  '  '  'to  b  factors 

_  ^a+a  +  a+  ' ' '  .  .  .  to  6  terms  of  the  exponent 

IV.  Law  of  Evolution,  or  extraction  of  roots, 
Law  I  may  be  stated  more  completely  thus : 


x""  'X^  'X'  "  .  =  2:^+^  +  «+--'. 

Law 

Ill  includes  the  more  general  forms 

(a) 

Qx^yhy  —  ^ac^ftc^ 

(^) 

((xf'yy ' .  .  =  x^^''-'. 

When  Laws  I,  II,  and  III  were  used  in  previous  work 
in  multiplication  and  division,  we  always  assumed  that  a 
and  b  were  positive  integers  and,  in  Law  II,  that  a  was 
greater  than  b.  In  the  work  on  radicals  (see  "First  Course 
in  Algebra,"  Revised  Edition,  pp.  251-252)  the  meaning 
of  an  exponent  was  extended  so  as  to  include  fractional 
exponents,  as  defined  by  Law  IV.  Though  Laws  I-IV 
have  thus  far  been  restricted  to  positive  integers  and  frac- 
tions, they  hold,  nevertheless,  for  any  rational  values  of  a 
and  b.  This  fact  will  be  assumed  without  proof.  We  shall 
now  explain  the  meaning  which,  according  to  these  laws, 
must  be  given  to  a  zero  or  to  a  negative  exponent. 


EXPONENTS  93 

52.  Meaning  of  zero  as  an  exponent.    From  Law  II, 

But  a^^a^  =  -  =  l. 

Therefore  ;c**  =  1. 

More  generally,      2f^  -h  sf'  =  af-'^  =  x^, 

and,  as  before,  aP  =1. 

That  is,  any  number  (except  zero)  whose  exponent  is  zero  is  equal 
to  1.  Hence  4^  =  (f )^  =  (—  6)^,  for  each  equals  1.  Again,  if  x  is  not 
zero,  (5  xy  =  1,  and  if  a;  is  not  1,  (x^  —  2  x  +  1)*^  =  1. 

53.  Meaning  of  a  negative  exponent.    From  Law  II, 


a' 


d> 


Obviously,  a^  _j_  ^5  _ 

Therefore  ar'^  is  another  way  of  writing  —  • 

Then  4-3^1_  ^  .' 

43      64 

Also  a"  t  z=  —  z=  — ^=^ . 

1 

In  general  terms,        x~^  =  — 

Consequently  =  —  =  Jf°. 

Similarly,  we  obtain  the  more  general  results 
hx~  "  =  —     and     =  hxf^. 


94  SECOND  COURSE  IN  ALGEBRA 

Therefore,  Any  factor  of  the  numerator  of  a  fraction 
may  he  omitted  from  the  numerator  and  written  as  a  factor 
of  the  denominator,  and  vice  versa,  if  the  sign  of  the  exponent 
of  the  factor  he  changed. 

It  follows  that  any  expression  involving  negative  exponents  may 
be  written  as  an  expression  involving  only  positive  exponents.  That 
is  to  say,  negative  exponents  are  not  a  mathematical  necessity  but 
merely  a  convenience.  The  extension  of  the  laws  of  exponents 
which  brings  with  it  the  zero  exponent  and  the  negative  exponent 
is  an  illustration  of  what  is  called  the  Law  of  Permanence  of  Form. 

It  is  to  be  understood  that  the  part  of  the  rule  for 
muItipHcation  (p.  5)  and  of  the  rule  for  division  (p.  7), 
which  determines  the  exponents  in  the  product  or  m  the 
quotient,  applies  to  all  numbers,  whether  positive  or  nega- 
tive, integral,  fractional,  or  literal.  Hence  those  rules  need 
not  be  restated  here. 

ORAL  EXERCISES 
Fractional,  Negative,  and  Zero  Exponents 

Simplify : 


1. 

xKcci 

10.  x^-^x-\ 

19. 

{a^-a^).aK 

2. 

cc^^l^ 

11.  x^  -x-^. 

20. 

{al^a^j^l)a\. 

3. 

x^.xl 

12.  x^  'X-^. 

21. 

^8a-l  ^  x'^-'2a 

4. 

x^.xi. 

13.  x:'-^x-\ 

22. 

x«  -  x\ 

5. 

X"  -T-  X^. 

14.  x^-i-x-\ 

23. 

x2«-i.xi 

6. 

x'  .  x\ 

15.  {bxy-nx'. 

24. 

Vx  'X. 

x^.x. 

16.  lx''x\ 

25. 

^X  -h  x\ 

7. 

8. 

x" .  x°. 

17.  x^  'X^. 

26. 

vs.^i. 

9. 

aj«  'X. 

18.  ai  .  a^  '  a^. 

27. 

Vi*--  Vi. 

EXPONENTS  95 


28.  ^x  .  Vx\  33.   {x'^y  .  x\  37^  ^2  _^  J_ 

29.  V^.  V^.  34.  (x^a-x-''-. 


X 

1 


30.  a^^-^-x^-^.  35.  a^^^^-  38.  ^   ^  :  ^_, 

x~^ 

31.  x2«-^--cc^-2«.  ^       •         39.  ici-^*-T-x-2«. 

32.  (x2)3  .  x\  ^^'  ^^"  ^  ^ '  40.  ax-^  -^  a-^x\ 

Kead  the  following  with  positive  exponents  and  simplify  the 
results : 


41. 
42. 

2a-\ 

49. 

4c« 
xy-^ 

54. 

10- ^a 
bc^ 

43. 
44. 

Sab--\ 

lxhj-\ 

50. 

y-' 

55. 

45. 
46. 

x~'^y~^z. 

51. 

5-\aby 

10- •■^^»'^ 

56. 

4-V-V- 

47. 

3 

52. 

■3a^b-^c 

57. 

2  s-' 

48. 

4:X 

53. 

Vlxhj-' 
2yx-^ 

58. 

2e-:    ^ 

Arrange  terms  so  that  the  exponents  of  one  letter  occur  in 
the  descending  order : 

59.  a^-\-a-'-ea-{-Sa''. 

60.  a^-]-a~^^~-a^ -i-a  +  Sa'. 

61.  a  +  a^  +  2  +  a-^  i-a-\ 

62.  a^  +  a~^  +  a~^  -\- a -j- S. 

63.  a^  +  -^  +  «  +  -  +  6^'. 

67.  Arrange  the  polynomials  in  Exercises  17  and  21  on 
page  97  in  descending  order  with  respect  to  the  letter  a,  and 
Exercises  22  (p.  97)  and  28  (p.  98)  with  respect  to  x. 


64. 

a^               a 

65. 

••4.+^& 

66. 

--S-?-¥-^ 

96  SECOND  COURSE  IN  ALGEBRA 

EXERCISES  IN  MULTIPLICATION 
Perform  the  indicated  multiplication  and  simplify  : 

1.  {x  +  x^-^x-^)2xi.  13.  (e^  +  e-y. 

2.  \a^x-[-ax^  —  a^x  ^)ax^.  ^  ^ 

3.  (x^  —  a^)x^a^.  ^  /I       1\ 

4.  (x^-5ax  +  6a^)Jx-k      ^^'  ^^'" +.^"H^  "  ^)* 

5.  (x^  +  y^jx^yK  ^  ^  ^ 
/I         IX  /   1          tx              18.  (3a-^  +  2a*)'. 

19.  (x-i-3x-2ir-2)2. 

7.  (cc^  +  y'^) (cc^  —  2/^).  "         /      1       ^    1       ^    3\2 

^  ^^  -^  ^  20.  U-^4-2a;^-3x^). 

8.  (a-2  +  3)(a-2-5). 

9.  (.--3.«)(.^-2.).       ^1-  (-*  +  -^+l)U^--^+l)- 

10.  (a-i  -  a)l  ^  o        , 

11.  (6i«-2«-y.  23.  {x^-(^){x^-\-a^-J^aS). 

12.  (a- 1  _  2  a  +  3  a- 2)2.  24.  (a^ -  a^cc  +  4  ic^) (a^  +  2 ic). 

25.  (x^  +  2  ?/^) (x3-  -  2 x^?/^  +  4  2/^). 

26.  (a  +  a^^>^  +  ^)U  +  ^^  -  f^^^'^). 

27.  (a*  -  <i*^»*  +  h^)  {h^  +  «^  +  o^^^'^). 

28.  {x  4-  xV^  +  y~^)^y~^  +.^  -  ^^2/"^)- 

29.  (^/^^-5Va)(>/^^-5V^). 

Hint.   This  is  best  written  (ai  —  6  a^)  (a^  —  6  a^). 

30.  (5  V^  +  -v^^  -  a  -v/^'f. 

31.  W-^^n^-v;^). 


5. 


EXPONENTS  97 

EXERCISES  m  DIVISION 

Perform  the  indicated  division  and  simplify : 

1.  x*^x\  6.  (x«- 2x2^-1  + 3 ic^«-2)-f-x2«-i. 

2.x^^x^.  7.   (6  67.3+*^-9a^™-2  +  12a2-n^)_^3^»-2^ 

3,  x^  ^  x^.  8.   (x-y)-^  {x^  —  y^). 

4.  axi-i-Jx^.  9.  (x  +  y)^{x^  +  yi). 
ax-a^x\              10.   (a^-8  2/)-(aj*-2  2/^). 

a^a:^  11.  (16a^^-A096f)-i-{2x^-^Syi), 

12.  (at  +  Z,)-^(V^-^^). 

13.  (a^  +  c»6-i  +  ^*-2)^(^^  _  Jf^-h  +  ^-i). 

14.  (e2^4-e-2^  +  2)--(e-^  +  e^). 

16.  (6  +  e-^^  +  e^o;  _  4  ^-20;  _^  4  e^^^)-- (e^  -  e"^). 

17.  (a  +  2  Z^  +  2  ah^  +  a*^»*)  -  («^*  +  2  5^). 

18.  (m'  -  7m-2  +  m"*  +  7  m^  +  8)  -  (5  -  7/^-^  +  m^). 

19.  (9a;^*^-^-.T3«-2_^2cc2«-i-)_j_^2ic'^-i  +  3ic2"-2). 

20.  (I6 £c  -  8 y~^  -\-x~^y-2x~  V) ^ (^~ ^y~^  -S xhj- 1). 

21.  (40  ah  -  2^a-h^  -  16  aH^)  -  (-  4  a'^V^  +  5  a^h-^) 

22.  (2:r-2«  -  28.T-«  +  33  +  lla^^^  +  38a:«  +  cc^^) 

-^(4  +  a;«-2cc-«). 

Divide : 

23.  a^  -  ah^  +  a^^  -  h^  by  a^  -  5^ 

24.  3£c-i°  +  x«-4a^-«by  2ir-2  +  a^2_^3a^-« 


98  SECOND  COUESE  IN  ALGEBRA 

25.  x^  —  y^  by  \_{x^  —  y^)-i-{x^^  +  2/^)]. 

26.  9m  +  4m-i-13  by  3w^-5  +  2m~i 

27.  ^2a  ^  4^-2a  _  29  by  a!«  _  2cc-«  -  5. 

28.  9x2«  +  25£c-*«  -  19x-«  by  5a:-2«  +  Sic*^  -  Ix 


29    (^  +  ?I^ 
?9-  '  8  ^    64 


^  [/^^  -f  ^"j  (64  a;  -  96  x^y^  +  144  y)l. 


Note.  To  us,  who  use  the  notation  of  exponents  every  day,  it 
seems  so  simple  and  natural  a  method  of  expressing  the  product  of 
several  factors  that  it  is  difficult  to  understand  why  such  a  long 
time  was  necessary  to  develop  it.  But  here,  as  in  many  other  in- 
stances, it  required  a  great  man  to  discover  what  to  us  seems  the 
most  obvious  relation.  The  man  who  brought  the  notation  of 
exponents  to  its  modern  form  was  John  Wallis  (1616-1703),  an 
Englishman. 

Though  the  idea  of  using  negative  and  fractional  exponents  had 
occurred  to  writers  before  Wallis,  it  was  he  who  showed  their  natu- 
ralness, and  who  introduced  them  permanently.  He  also  was  the 
first  to  use  the  ordinary  sign  oo  to  denote  infinity. 

MISCELLANEOUS  EXERCISES  ON  EXPONENTS 

Find  the  numerical  value  of  the  following : 


1.  3-^ 

10.  1- 

18.  16"^. 

2.  4-^ 

3° 

19.  8-1 

3.  2-*  •  3^ 

20.  16-1 

4.  2--.  3-". 

12.  5.2°- 

.(5  .  2)^ 

21.  25  H 

5.  7  .  T'' .  0. 

,o    4-^3- 

-2 

22.  (-  64)-*. 

6.  a)-^ 

13.       ,_. 

~* 

23.  (-  32)i 

7.  (|)-^.4o. 

14.  32-1 

24.  (32)-*. 

8.  (i)-^  •(¥)-'• 

15.  0^ .  h\ 

25.  (-125)- i 

2 

16.  4"i 

26.   ■v^27-^ 

*-3-- 

17.  8-^. 

27.   -^8-^. 

EXPONENTS 


99 


ay- 


22      23 


28.  {^/^sy. 

30.  (I)" I  ^^^'^-  ^r-7. — ::r-.=   .  '  .  ^  etc. 

31.  (.04)1 

32.  (.027)" i 

33.  (.064)- 1 

34.  (.00032)1 

Write  with  positive  exponents  and  simplify  the  results 


36. 

2-1 

2- 

-2 2-3 

TTttx 

rT. 

2-1 

2-^-2-3 

37. 

3- 

-2  2-2 

3- 

■1-2-1 

38. 

2- 

■1  +  3-1 

2- 

-3_^3-8 

QO 

3^ 

-3  9-3 

40. 


Hint. 


44. 


2 


5-2      J.  _  1_ 

-2 

46. 


etc. 


41. 
42. 
43. 


+  ^'- 


45. 


a-%- 


7.-4 


-hb- 


47. 


a-^-b-' 


a-^  +  b- 

-1 

a 

a-'~  -  b- 

-2 

Bse" 

s--  +  e- 

-2 

18     '~' 

+  ^»-l 

*®-   a-3 

+  Z»-3 

-27-1 

-3-^ 


"V^rite  without  a  denominator  and  simplify  the  results : 


50. 
51. 
52. 
53. 


2xy 
b'  ' 

4  6-c-^ 
2-is-' 


54. 
55. 
56. 
57. 


12  a%^ 

4.xf  ' 
Ix-hf 

2-y  * 

c{x-yf 


58 


59.    ^(^-?/)-^ 

^6'iC  (iC  —  y) 

42m-«?i2"* 


60. 
61. 


r~V(s  —  r)* 


100  SECOND  COURSE  IN  ALGEBRA 

Simplify : 

62.  (28/.  71.  (x-^yK  80.  (a8)2^  •  (a2)8^. 

63.  (28)-2.  72.  (Sx-'^y.  81.  (a^+i)^  •  (ai-^)^ 

64.  (2-')-\  73.  (5(^^)-^  82.  (««)^  +  ^-(a'')^-^ 

65.  [(f)-^^  74.  (0-^)^.  83.  (a%y  •  (W)^ 

66.  W-  75.  (5o.2«.38)i  8^.  (.^-.->3. 

67.  U^j.  76.  (25a^^«)-^ 

68.  {x^)\  77.  (2a?y  .  8-4^. 


^2^8\2a? 


yn       C2  71 

88.  ^^-;:2_  =  2 


86.  2'^.  42  =  2?. 

87.  2"  .4^  =  2?. 

69.  (xy\  78.  (6i2^^)«(a  +  Z.).  2 

88    — 

70.  (x-«)-2.  79.  (a^)^'^  .  a^^.  *       43^^ 

89.  2''  .4^+^^  2'' =  2?. 

4.TC  +  1                        J^^+1 
Oft        : _J_  5   —   9 

2"('4'''~^V*  *   /4»  +  iyi-i  "^ 
Solve  for  w: 

^^  -95.  81.27^^=(9")2. 

92.  3«.9^  =  812.  g,„ 

93.  9'*  .  3«  =  27^\  ^^-  (^^'')''  ==  (l25y^' 

94.  2^'^"^^  •  4"  +  '^  =  8^".  97.  2^'"''^^  •  4^^+^  =  ('8"')". 

Solve  for  x : 

98.  a-^  =  8.  103.  x-^  =  4.  108.  ^x~^  =  2. 

99.  x-^  =  5.  104.  x^  =  2.  109.  (cc"*)"*  =  49. 

100.  x~^=25e.         105.  £c^  =  16.  110.  (aici)"*  =  27. 

101.  x"i  =  2.  106.  ic"^  =  32.  V^  _  \/25 

102.  x^  =  4.  107.  x^  =  343.  *  ^x^      -^ 


CHAPTER  VII 
SQUARE  ROOT 

54.  Square  root.  The  square  root  of  any  number  is  one 
of  the  two  equal  factors  whose  product  is  the  number. 

From  the  law  of  signs  m  multiplication  it  follows  that 
Every  positive  number  or  algebraic  expression  has  two  square 
roots  which  have  the  same  absolute  value  but  opposite  signs, 

55.  Square  root  of  a  monomial.  For  extracting  the 
square  roots  of  any  monomial  we  have  the 

Rule.  Write  the  square  root  of  the  numerical  coefficient 
preceded  by  the  sign  ±  and  followed  by  all  the  letters  of  the 
monomial^  giving  to  each  letter  an  exponent  equal  to  one  half 
its  exponent  in  the  monomial. 

A  rule  similar  to  the  preceding  one  holds,  for  the  fourth 
root,  the  sixth  root,  and  other  even  roots. 

56.  Cube  root.  The  cube  root  of  any  number  is  one  of 
the  three  equa,l  factors  whose  product  is  the  number. 

For  extracting  the  cube  root  of  a  monomial  we  have  the 

Rule.  Write  the  cube  root  of  the  numerical  coefficient  fol- 
lowed by  all  the  letters  of  the  monomial^  giving  to  each  letter 
an  exponent  equal  to  one  third  of  its  exponent  in  the  monomial. 

A  rule  similar  to  the  preceding  one  holds  for  the  fifth 
root,  the  seventh  root,  and  other  odd  roots. 

57.  Principal  root.  For  a  given  index  the  principal  root 
of  a  number  is  its  one  real  root  if  it  has  but  one,  or  its 
positive  real  root  if  it  has  two  real  roots  of  that  index. 

101 


102  sircc»:^rD  course  in  algebra 

Then  the  principal  square  root  of  9  is  +  3 ;  the  principal  fourth 
root  of  16  is  +  2,  not  —  2.  The  square  root  of  —  4  or  —  9  is  not 
real ;  such  numbers  have  no  principal  square  root. 

The  principal  cube  root  of  8  is  2,  of  —  27  is  —  3.  The  principal 
fifth  root  of  32  is  +2,  of  -  32  is  -  2. 

Every  number  has  more  than  one  root  of  given  odd  index. 
The  number  8,  for  example,  has  two  other  cube  roots  besides 
its  principal  cube  root  2.  What  they  are  and  how  they  are 
obtained  will  be  made  clear  in  the  chapter  on  Imaginaries, 
where  the  consideration  of  the  square  roots  of  negative 
numbers  will  also  be  taken  up. 

Unless  otherwise  specified,  only  the  principal  odd  root 
of  a  number  will  be  considered. 

ORAL  EXERCISES 

Find  the  principal  square  root  of  the  following : 


1.  16. 

2.  25. 

3.  4.a\ 

5.  36  a«. 

6.  49  t\ 

7.  64 1\^ 

9. 
10. 

h                 xJ" 
^'          13.  x-\ 

15.  9a;V*. 

16.  4x\ 

17.  x^. 

4.  9  a*. 

8.  81a?«. 

11. 

18.  x^. 

Find  the 

principal  cube  root  of  the  following 

19.  8. 

24.  27  rr*. 

29.  -125a«. 

34.  343  a«. 

20.  27. 

25.  64  x\ 

30.  -  27  a\ 

35.  -512a«. 

21.  64. 

26.  -8. 

31.  -64a\ 

36.  -  27  a^. 

22.  8  a". 

27.  -27. 

32.  -%a\ 

37.  -^x-\ 

23.  8a«. 

28.  -  64. 

33.  216  a\ 

38.  -27ic-«. 

Find  the  principal  fourth  root  of  the  following : 

39.  16.  42.  a\  45.  16  x\  48.  x^, 

40.  81.  43.  x\  46.  625  a^  49.  16  a;-*. 

41.  266.  44.  x"^.  47.  16x-\  50.  81  x"^. 


SQUARE  BOOT  103 

Give  the  principal  root  and  one  other  root  for  the  following : 

51.  The  fourth  root  of  81;  of  w"  \  of  x~^. 

52.  The  sixth  root  of  64;  of  a«;  of  x-\ 

53.  What  is  the  sign  of  the  principal  odd-Yoo^,  of  a  positive 
number  ?    The  principal  odd  root  of  a  negative  number  ? 

54.  What  is  the  sign  of  the  principal  even  root  of  a  positive 
number  ? 

55.  State  the  rule  for  extracting  the  fourth  root  of  a  mono- 
mial. 

56.  State  the  rule  for  extracting  the  fifth  root  of  a  monomial. 

57.  Can  one  obtain  the  fifth  root  of  a  monomial  by  extract- 
ing the  square  root  of  its  cube  root  ?  by  extracting  the  cube 
root  of  its  square  root  ?    Explain. 

58.  Square  root  of  polynomials.  Extractmg  the  square 
root  of  a  number  is  essentially  an  undoing  of  the  work 
of  multiplication.  The  square  of  any  polynomial*  may  be 
represented  by 

Qi^t^uy'  =  h^  +  2]it  +  t'^^-2hu  +  2  tu  4-  w2 

A  little  study  of  this  last  form  and  a  comparison  with 
the  example  which  follows  will  make  clear  the  reason  for 
each  step  of  the  process. 

EXAMPLE 

J^ 

First  trial  divisor,  2  h 


+  2]it-\-t^ =(2A+  0^ 


First  complete  divisor,  2  A  +  ^ 

Second  trial  divisor,  2^  +  2^  |+2AM  +  2m+u2  =  (2^  +  2«+w)u 

Second  complete  divisor,  2 ^  +  2 « +  m \+2Tiu-\-2tu-\-u'^  =  {2h-\-2t-\- u)u 

Therefore  th^e  required  square  roots  are  ±  (h  -{■  t  ■\-  u). 


104 


SECOND  COURSE  IN  ALGEBRA 


EXERCISES 

Extract  the  square  root  of  the  following : 

1.  x^-{-4.x^-2x^-12x-\-9. 

2.  a^-10d^-4.a^-\-25a^-{-20a-\-4.. 

3.  x^  +  4lx^  +  16-4.x^-{-Sx^-16x. 

4.  t^  ^  4:t^  -{-  9  +  4.t'  -  6 1^  -  12  t\ 

5.  4  a*  +  9  aH^  +  t^  +*12  aH  -  4  aH""  -  6  a^^- 

6.  4  a«  +  12  c^*  -  7  -  24  c*"*  +  16  «"«. 

7.  49  c-«  -  28  c-4  +  74  c-^  -20  +  25  e". 

8.  9x2  +  4a^  +  l  +  12ir*4-6x  +  4£A 

9.  9i»*-6a^^  +  a;^-66x^  +  22ic2_^121:r. 

10.  25x^-10x''-\-90xi  +  x-lSxi-j-Slxk 

11.  16  772.-^- 


12.  -^  +  7  +  /-7,  + 


-*  +  104 m  -  26 m*  +  169  m«  +  m-\ 
3  2/ 


iC 


Solution.    Arranging  terms  in  descending  powers  of  x  and  apply- 
ing the  method  of  page  103,  we  obtain  the  following : 


(t) 


4x2      12 

y^       y 

4a:2 


X  4:X^ 


_  +  3-^ 


r 


ff-)= 


4x 

y 
i^  +  3 

4a: 


12  X 

y 
y 


"(t-)^ 


+  6 


i£  +  „_JL 

V  2  a; 


2_^  +  JC 
a:        4ar2 


-2 


a:        4 


a:2      \?/   ^  2a:/\     2x1 


Therefore  the  square  roots  are  ±(—  +  3  —  ^j- 


SQUARE  ROOT  105 
13.  x«  +  6a;'^-^  +  2x  +  -• 
15.  ^ ^g     -2a;+     3     +^4- 

4       9  o 

17.  ^a'  -  2a'  -\-7^a'  -  -^^a  -{-  ^K 

18.  9c*-12c^  +  4c2--  +  i  +  6. 

19.  ^'-18  +  -,  +  9(^2-2  +  2a2. 
9  a^ 

20.  ^ 1 +  -^  +  4i. 

21-  I^  +  ^  +  ^'^'  +  2^2  +  2  +  2^2. 

x^      a^        X         a 

■  g^         2a2      117       40       16 

**'4c*^     ^25^*        c^^Sac       5^2 

^^-  9c2"^    a^    "*"3c"^    g    "^3' 

J[__3£      9^      _a^_6^      J^ 
4a*       a«  ■*"   a"-   "^25:^2        5    "^5a;' 

Find  the  first  four  terms  in  the  square  root  of  the  following : 
27.  l  +  2a;.  28.  ^'^-a?. 

KE 


106  SECOND  COUKSE  IN  ALGEBRA 

59.  Square  root  of  arithmetical  numbers.  The  abbrevi- 
ated process  of  extracting  the  square  roots  of  an  arithmetical 
number  is  as  follows : 

7^32^67^89 1 2706.8  + 
4 
47 


332 
329 


5406 
54128 


36789 
32436 


435300 
433024 


2276 


Therefore  the  square  roots  of  7326789  are  ±2706.8+. 

It  follows  from  the  preceding  example  that  the  work  of 
extracting  the  positive  square  root  of  a  number  may  be  a 
never-ending  process.  The  number  7,326,789  has  no  exact 
square  root,  and  no  matter  how  far  the  work  is  carried, 
there  is  no  final  digit.  As  the  work  stands  we  know  that 
the  required  root  lies  between  2706.8  and  2706.9. 

The  method  just  illustrated  for  extracting  the  positive 
square  root  of  a  number  is  the  one  eommonly  used.  For 
it  we  have  the 

Rule.  Begin  at  the  decimal  point  and  point  off  as  many 
periods  of  two  digits  each  as  possible :  to  the  left  if  the  num- 
ber is  an  integer^  to  the  right  if  it  is  a  decimal^' to  both  the  left 
and  the  right  if  the  number  is  part  integral  and  part  decimal. 

Find  the  greatest  integer  whose  square  is  equal  to  or  less 
than  the  left-hand  period  and  write  this  integer  for  the  first 
digit  of  the  root. 

Square  the  first  digit  of  the  root^  subtract  its  square  from 
the  first  period  and  annex  the  second  period  to  the  remainder. 


SQUARE  KOOT  107 

Double  the  part  of  the  root  already  founds  for  a  trial  divisor, 
divide  it  into  the  remainder  (omitting  from  the  latter  the  right- 
hand  digit^,  and  write  the  integral  part  of  the  quotient  as  the 
next  digit  of  the  root. 

Annex  the  root  digit  just  found  to  the  trial  divisor  to  make 
the  complete  divisor,  multiply  the  complete  divisor  hy  this  root 
digit,  subtract  the  result  from  the  dividend,  and  annex  to  the 
remainder  the  next  period,  thus  making  a  new  dividend. 

Double  the  part  of  the  root  already  found,  for  a  new  trial 
divisor,  and  proceed  as  before  until  the  desired  number  of  digits 
of  the  root  have  been  found. 

After  extracting  the  square  root  of  a  number  involving 
decimals,  point  off  one  decimal  place  in  the  root  for  every 
decimal  period  in  the  number. 

Check.  If  the  root  is  exact,  square  it.  The  result  should 
be  the  original  number.  If  the  root  is  inexact,  square  it  and 
add  to  this  result  the  remainder.  The  sum  should  be  the 
original  number. 

EXERCISES 

Find  the  positive  square  root  of  the  following : 

1.  6241.     '  5.  53.29.  9.  2,932,900. 

2.  9216.  6.  1.4641.  10.  7,049,025. 

3.  15,129.  7.  216.09.  11.  3.9601. 

4.  56,169.  8.  988,036.  12.  .0061504. 

Find  to  three  decimal  places  the  square  root  of  the  following: 

13.  7.  15.  .01235.  17.  f.  19.  ^K  21.  23^\. 

14.  .63.        16.  .96384.  18.  4|.        20.  |.  22.  89i. 

23.  Find  the  hypotenuse  of  a  right  triangle  whose  legs  are 
136  and  273  respectively. 


108  SECOND  COUESE  IN  ALGEBRA 

24.  A  baseball  diamond  is  a  square  90  feet  on  each  side. 
Find  the  distance  from  the  home  plate  to  second  base,  correct 
to  .01  of  a  foot. 

25.  The  hypotenuse  of  a  right  ttiangle  is  207  feet,  and  one 
leg  is  83  feet.   Find  the  other  leg,  correct  to  .01  of  a  foot. 

26.  The  hypotenuse  and  one  leg  of  a  right  triangle  are 
respectively  5471  and  4059.    Find  the  other  leg. 

27.  The  side  of  an  equilateral  triangle  is  17  inches.  Find 
its  altitude,  correct  to  .1  of  an  inch. 

28.  Find  the  side  of  an  equilateral  triangle  whose  altitude 
is  15  inches,  correct  to  .001  of  an  inch. 

Fact  from  Geometry.  If  a,  ^,  and  c  represent  the  sides  of  a 
triangle  and  s  equals  one  half  of  a  +  &  +  c,  the  area  of  the 
triangle  equals  Vs(s  —  a){s  —  b)  (s  —  c). 

29.  Find  the  area  of  a  triangle  whose  sides  are  12,  27,  and 
35  inches  respectively,  correct  to  .001  of  a  square  foot. 

30.  By  the  method  of  Exercise  29  find  to  .01  of  a  square 
inch  the  area  of  a  triangle  each  side  of  which  is  22  inches. 

31.  Find  to  two  decimals  the  sum  of  all  of  the  diagonal 
lines  that  can  be  drawn  on  the  faces  of  a  cube  whose  edge  is 
11  inches. 

32.  Find  to  two  decimals  the  radius  of  a  circle  whose  area 
is  70  square  feet. 

33.  Find  to  two  decimals  the  diagonal  of  a  room  whose 
dimensions  in  feet  are  15,  22,  and  "28. 

34.  Find  to  two  decimals  the  diagonal  of  a  cube  whose  edge 
is  8  feet. 

35.  A  room  is*  24  feet  by  40  feet  by  14  feet.  What  is  the 
length  of  the  shortest  broken  line  from  one  lower  corner  to 
the  diagonally  opposite  upper  corner,  the  line  to  be  on  the 
walls  or  the  floor,  but  not  through  the  air  ? 


CHAPTER  VIII 

RADICALS 

60.  Radical.  A  radical  is  an  indicated  root  of  an  algebraic 
or  aritnmetical  expression. 

Thus  V9,  V5,  V2x,  and  Va;^  —  a;  —  12  are  radicals. 

61.  Index.  The  small  figure  like  the  3  in  v^7  is  called 
the  index  of  the  radical. 

The  index  determines  the  order  of  the  radical  and 
indicatefcj  the  root  to  be   extracted. 

For  square  root  the  index  is  usually  omitted.  Thus  V 2  and  ViS 
mean  v2  and  vl8  respectively. 

62.  Radicand.  The  radicand  is  the  number  or  expression 
under  the  radical  sign.  In  V5  and  \^2ax  the  respective 
radicands  are  5  and  2  ax. 

63.  Fractional  exponents.  Radical  expressions  may  be 
written  in  either  of  two  ways :  with  radical  signs  or  with 
fractional  exponents. 

Thus  v/5  and  5^  have  the  same  meaning,  and  Va^  equals  as:,  etc. 

64.  Rational  numbers.  A  rational  number  is  a  positive  or 
a  negative  integer  or  any  number  which  can  be  expressed 
as  the  quotient  of  two  such  iutegers. 

Thus  7,  —  6,  |,  or  2.871  are  rational  numbers. 

65.  Irrational  numbers.  Any  real  number  which  is  not 
rational  is  irrational.   (See  section  67.) 

109 


110  SECOKB  COURSE  IN  ALGEBRA 

If  a  number  under  a  radical  sign  is  such  that  the  root 
indicated  cannot  be  exactly  obtained,  the  radical  represents 
an  irrational  number. 

For  example,  V?  and  Vi  are  irrational.  Approximate  values  for 
these  are  given  on  page  274. 

A  repeating  decimal,  though  endless,  is  not  an  irrational 
number, '  for  any  repeating  decimal  can  be  expressed  as  a 
common  fraction,  and  is  therefore  rational. 

Thus  the  repeating  decimal  .272727  •  •  •  is  not  irrational,  as  it 
exactly  equals  ^^.   Similarly,  .2857142857142  •  •  •  exactly  equals  f ,  etc. 

Note.  The  number  f  reduced  to  a  decimal  repeats  the  digits  in 
groups  of  six  each,  and  the  mere  fact  that  a  decimal  does  so  repeat 
is  proof  that  it  is  a  rational  number.  On  the  other  hand,  the  num- 
ber TT  is  known  to  be  irrational,  and  its  value  has  been  computed  to 
707  decimals,  showing,  of  course,  no  repetition.  The  fact  that  it 
does  not  repeat  in  700  digits  is,  however,  no  proof  that  tt  is  irrational, 
for  decimals  with  even  more  than  that  many  digits  do  repeat.  For 
example,  the  fraction  WgV^  equals  the  decimal  1.29 +  ,  which  repeats 
in  groups  of  7698  digits  each. 

66.  Imaginary.  An  indicated  square  root  of  a  negative 
number  is  called  an  imaginary  number. 


Thus  V—  4,  V—  7,  and  V— 12  are  imaginary  numbers.  And 
3  +  V^^  is  also  imaginary,  though,  as  will  be  seen  later  (Chapter  XII), 
such  numbers  are  better  called  complex  numbers. 

67.  Classification  of  numbers.  All  the  numbers  of  alge- 
bra then  may  be  placed  in  one  or  the  other  of  two  classes : 
real  numbers  and  imaginary  numbers. 

Real  numbers,  as  we  have  seen,  are  of  two  kinds,  rational 
numbers  and  irrational  numbers. 

68.  Surd.  A  surd  is  an  irrational  number  in  which  the 
radicand  is  rational. 

Thus  Va,  V^,  etc.,  are  surds.    But  v  2  +  Vi}  and  Vtt  are  not  surds. 


EADICALS  111 

69.  The  algebraic  sign  of  a  radical.  The  square  root  of 
4  is  both  +  2  and  —  2.  The  symbol  Vl,  however,  sig- 
nifies only  +  2,  the  principal  root  (section  57).  Similarly, 
^81  is  +  3  and  V64  is  +  2^  But  -V9  is  -  3,  and  -^l6 
is  —  2.  The  symbol  ±  V25  denotes  both  +  5  and  —  5. 
Further,  +^27- +3,  -^27  =  - 3,  and  -^327== +3. 

The  foregoing  remarks  apply  also  to  fractional  expo- 
nents. Thus  4^  =  +  2  only,  and  81^  =  +  3  only,  etc.  It 
should  be  noted  that  these  statements  really  define  the 
meaning  of  such  symbols  as  V~?  ^?  v^  »  etc.  Such  an  un- 
derstanding as  this  avoids  all  the  ambiguity  which  would 
arise  if  Vl6  meant  both  +  4  and  —  4.  The  distinctions 
here  made  are  especially  needed  in  radical  equations. 


ORAL  EXERCISES 

Find  the  numerical  value  of  the  following : 

1. -Vi. 

7.  -^81. 

13.  8* 

19.  16^ 

2.  -V9. 
/ — 

8.   v^32. 

14.  36t 

20.   (i)^. 

3.   Vl6. 

9.  -^-125. 

15.  49^. 

21.   (f)*. 

4.   Vs. 

10.  --v^-32. 

16.  8^. 

22.  (-  64)*- 

5.    ^-8. 

11.   ^64. 

17.  16*. 

23.   (49)1 

6.  -^/i6. 

12.  -^625. 

18.  27^. 

24.  (121)1 

Bead  in  radical  form  : 

25.  x\ 

30.   Srx^. 

35. 

a(a-l)^. 

26.  x^. 

31.  4.ax^. 

36. 

2c(2x-3)i. 

27.  (at)\ 

32.  2aV. 

37. 

x^'k 

28.   (3t)i 

33.  5ux^. 

38. 

2ixk 

29.  St^. 

34.  4Ai 

39. 

a  2  71 

xH'^. 

112  SECOND  COURSE  IN  ALGEBRA 

Read  with  fractional  exponents : 

40.  V^^  43.   V'^K  46.    -^{a  +  x)\ 

41.  V^«.  44.  2V^.  47.    -s/^^x-ay. 

42.  Va\  45.    ^v^.  48.    -^x'^y  -  1). 

49.  What  are  the  two  square  roots  of  36  ? 

50.  What  are  two  fourth  roots  of  16  ?  of  81  ?  of  625  ? 

51.  What  is  the  value  of  a/16  ?  of  "v^  ?  of  -v/625  ? 

52.  What  are  two  sixth  roots  of  64  ? 

53.  What  is  the  distinction  between  a  rational  number  and 
an  irrational  one  ? 

54.  Which  of  the  numbers  8,  |,  343,  VI,  Vs,  and  ir 
(it  =  3.14159  +  )  are  rational  ?    Which  are  irrational  ? 

55.  Give  a  geometrical  illustration  of  an  irrational  number 
by  means  of  a  right  triangle. 

56.  Is  a  radical  always  a  surd  ?    Illustrate. 

57.  Is  a  surd  always  a  radical  ?   Illustrate. 

58.  Distinguish  between  a  surd  and  a  radical. 

59.  Which  of  the  numbers  Vs,  VI,  ^27,  V^,  \/2  +  ^^3, 
and  V2  tt  are  surds  ?    Which  are  radicals  ? 

60.  What  is  the  principal  root  of  ±  Vi,  Vs,  and  V—  8  ? 

61.  Name  the  order  of  V6 ;  of  a^;  of  VS;  of  c*;  of  Vm^. 

62.  Give  an  example  of  (a)  a  real  number ;  (b)  an  imaginary 
number;  (c)  a  rational  number;  (^Z)  an  irrational  number; 
(e)  a  radical;  (/)  a  surd;  («7)  an  index;  (h)  a  radicand; 
(i)  the  principal  odd  root  of  a  positive  number ;  (j)  the  prin- 
cipal even  root  of  a  positive  number;  (k)  the  principal  odd 
root  of  a  negative  number. 

70.  Simplification  of  radicals.  The  form  of  a  radical 
expression  may  be  changed  without  altering  its  numerical 
value.    It  is  often  desirable  to  change  the  form  of  a  radical 


RADICALS  113 

so  that  its  numerical  value  can  be  computed  with  the  least 
possible  labor. 

The  simplification  of  a  radical  is  based  on  the  general 

A  radical  is  in  its  simplest  form  when  the  radicand 

/.  Is  integral. 

n.  Contains  no  rational  factor  raised  to  a  power  which 
is  equal  to,  or  greater  than,  the  order  of  the  radical. 

m.  Is  not  raised  to  a  power,  unless  the  exponent  of  the 
power  and  the  index  of  the  root  are  prime  to  each  other. 

For  the  meaning  of  I,  II,  and  III  study  carefully  the 

EXAMPLES 
Examples  of  I : 

-    1.  Vf  =  Vf  =  vT^  =  Vi^  =  i^- 

2.  6\4  =  6Vf  =  6vTr3  =  6.iV3  =  2V3. 
Examples  of  II : 


2.  5  ■\/24.x'  =5^Sx^-Sx^  =  5  V(2x)^  'Sx^  =  10a;  VS^. 

3.  Vl6  -  8  V2  =  \/4(4  -  2  V2)  =  2  V4  -  2  Vi 
Examples  of  III : 

1.  ■^  =  -v^  =  2*  =  2^  =  V2. 

2.  ^  =  ^2_3*  =  3^  =  -v^3. 
3.'   -\/'^^  =  Jb^  =  ah--=b\^. 


114  SECOND  COURSE  IN  ALGEBEA 

A  radical  of  the  second  order  is  simplified  by  the  use 
of  the 

Rule.  Separate  the  radicand  into  two  factors  one  of  which 
is  the  greatest  perfect  square  which  it  contains.  Tiien  take  the 
square  root  of  this  factor  and  write  it  as  the  coefficient  of  a 
radical  which  has  the  other  factor  as  radicand. 

If  the  original  radical  has  a  coefficient  other  than  the  num- 
ber i,  multiply  the  result  obtained  above  by  this  co&^cient. 

A  similar  rule  holds  for  simplifying  radicals  involving 
the  cube  root  and  roots  of  higher  orders. 

EXERCISES 

Simplify : 

1.  Vl8.  6.   V52.  11.  Vl92.  16.  S  Vu. 

2.  V20.  7.   V63.  12.  2V45.  17.   V^. 


3.   V28.  8.   V68.  .  13.   Vi6.  IS. 


ax". 


4.   Vii.  9.   V75.  14.  4V54.  19.    VaV 


5.  V5O.  10.  VT08.         15.  -v^.  20.  nVT^\ 

21.    V|. 

Solution.    V^  =  V5  =  V^~^  =  i V3. 

22.  x/f.  23.  Vf  24.  V\.  25.   V|. 

•  26.  ^. 

ya        _  

Solution.    ^ ft  =  ^/£  =  ^/i . «  =  -  V^. 
\a       \a2      \a2  a 


34.  6v 


27.   \  — 


28. 


«^         -35.  VWl?. 

IT  3|2^     32.  Vi_    36.  v.^^£(|y. 

\^«'  ^^-    NT*        33.   V-|.       37.   x/r+(|y. 


RADICALS  115 

38. 


44.  \/a^  +  aW3. 

45.  V16-8V3. 


39.  4'+gJ-  46.  V54-9Vi8. 

42.  Vrr^V!.  49.  ^/J»-|V3. 


Hint.    V4-8V3  = 
V4  (1  -  2  ^ 

43.  V36  +  I8V5. 

51.   -^. 

v/3). 

54. 
55. 
56. 

50. 

Soli 

ition. 

21  =  2^=  V2. 

52.  -\/4. 

53.  -Va^t 

«l4a« 

Express  entirely  under  the  radical  sign : 

59.  2V7.  63.  a-Va.  68.  e^Ve^  +  e"^. 


Solution.    2V7=         64.   2c^.        ^^     ,      .  in^O_ 


60.  3V5.  ee.x^^\      70.  (2^  +  1)^)^^ 

61.  4  Vs. 


^„    a  3f9  „^     .x-3a  3|      125 


62.2^8.  S^a^'  5        \(a5-3a)2 

Express  in  simplest  form  with  one  radical  sign : 
72.  VV2.  73.  VV^.  76.  VV^» 

Solution.     VV2=V2^  74.   VVa.  77.    V  Vo-. 


=  2i:=^.  75.   vWl  78.   V-?/^. 


116 


SECOND  COUESE  IN  ALGEBRA 


79. 


82.  VsVsVS.  85.   VVV^. 


80.  V-\/Sd^x. 

81.  V3V3. 


83. 


Vs. 


84.  2\/2V2. 


Find  by  the  formula  of  Exercise  28,  page  108,  the  areas  of 
the  triangles  whose  sides  are 

88.  6,  8,  and  10.  90.  33,  56,  and  65. 

89.  7,  24,  and  25.  91.  104,  153,  and  185. 

71.  Addition  and  subtraction  of  radicals.  Similar  radicals 
are  radicals  of  the  same  order,  with  radicands  which  are 
identical  or  which  can  be  made  so  by  simplification. 

The  sum  or  the  difference  of  similar  radicals  can  be 
expressed  as  one  term,  while  the  sum  or  difference  of 
dissimilar  radicals  can  only  be  mdicated. 


Simplify  and  collect : 

1.   V8  +  VI8. 
Solution.  Vs  +  V18  = 


EXERCISES 


>V^. 


2  V2  +  3  V2 

2.  VI  -f  3  V2. 

3.  V5O  +  V98-V32. 

4.  V12  +  5V75-2V27. 

5.  3VI8-V98  +  V128. 

6.  V75  +  3V147- V12. 

7.  2V54  +  V24-V96. 

8.  V45  -  V20  +  5  V245. 

9.  3V275  +  2V99-5V44. 
10.  -v/ie  +  "v/si  -  3 -v^. 


11. 
12. 
13. 
14. 
15. 
16. 
17. 

18. 

19. 
20. 


+  V375. 


V192  -  4  -V24  -, 

V54  +  -v/ie  -  VT28. 

^625  +  ^40  +  Vi35. 

loVf-V^  +  V^- 

3x/f  +  3Vi-2V^. 
a  Vac^—  -VA—  5  Va^. 

\Sa   .      13  X  lax 

[a  fa  |5x* 


EADICALS  117 


21.  v^32 x^  4-  a/1250 x  -  4^512 x  -  V2^. 

22.  V(a  ^cf-c  ^{a  +  c)2  +  2  c  V {a  +  cf. 

23.  -^(ct  _  cf  +  c  ^^2  -  2  ac  +  c^  +  (»  +  c)  -v/^ 


24.  \^--aJ-+aJ— -^— +  2--^J- 


■'±^'_9 


ac 

3y 


25.   ^^  _|-  ^(3  c^  +  9)  (ct  +  3)^  -  V8i  +  c,  Vg  -  4  -V3, 


26.  2  V9  a^  -  9  a^^*  -  3  V9  aV"  -  96*  +  V(a--^  -  V')  (a  +  &). 

36a^^-36^^^ 
a-hh 


27.  (a-^»)-J^  +  V25a^-25&^  +  ^A| 
^         ^    ya—h  a—h   > 


72.  Multiplication  of  real  radicals.  Real  radicals  of  the 
same  order  are  multiplied  as  follows : 

Example  1.    Multiply  2  -\/x  —  3  V«  —  4  ^ax  by  2  Vox. 

Solution.  2  Vx_—  3  V«  —  4  Vaa; 

2  Vaa: 

4  a:  V«  —  6  a  Vx  —  8  aa: 

Real  radicals  of  different  order  are  multiplied  as  follows  : 
Example  2.    Multiply  Vn  by  -^x. 
Solution.  Vn  =  n^  =  n5  =  "V^n^ 

"v/x  =  x^  =  xt  =  "v/^^. 
Then  Vn  •  -^a;  =  -v^  •  V^  =  xn^x^. 

The  method  of  multiplying  real  radicals  is  stated  in  the 

Rule,    If  necessary,  reduce  the  radicals  to  the  same  order. 

Find  the  products  of  the  coefficients  of  the  radicals  for  the 
CQefficient  of  the  radical  part  of  the  result. 

Multiple/  together  the  radicands  and  write  the  product  under 
the  common  radical  sign. 

Reduce  the  result  to  its  simplest  form. 

The  preceding  rule  does  not  hold  for  the  multiplication  of 
imaginary  numbers.    This  case  is  discussed  in  Chapter  XII. 


118  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 

Perform    the    indicated    multiplication    and    simplify    the 
products : 

1.  V3V27.  5.  VI- Vi- VI- 

2.  Vi2Vi8.  6.  (V3-Va)V2. 

3.  Vf  .  Vf .  7.  (  V2  -  3  VS)  V5. 

4.  V|->/t-  Vf-  8.  (V3-2V2)(V2-V3). 

9.  (V5-3V2)(2V5-V3). 
10.   (Va  — Vax)(Va +  2  Va^). 

11.  (V2  + V3)(V2- V3).  16.  (V3^-V2^)^ 

12.  (3V5- V2)(3V5  4-V2).  17.  {-^x  -  3)1 

13.  (V7-V5)(V7  +  V5).  18.  (2V3a:-l)'. 

14.  (2V3- V3)(2V3  + V3).  19.  (V^  -  V.X-  -  2)'. 

15.  (4V5  +  2V7)(4V5-2V7).      20.  3Vaj-3V4x-8, 

21.  (V5  -  V3  -  V2)(V5  4-  V3  -  V2). 

22.  (3  V2  +  2  V3  +  V30)(2  V2  +  2  V3  -  2  V5). 

2J.(;,-fV3)(2«  +  ^V5). 

Square : 


26.  V2  -  Va:  -  3.  29.    Va- -  3  -  Va:  +  3. 

27.  -v^  _  Va^4-4.  30.  2  Vx  —  3  V2  .r  +  1. 


28.   Vrr  -  3  +  Vic  +  5.  31.  3  V;r  -  1  -f-  2  vT 


a;. 


RADICALS  119 

Perform  the  indicated  multiplication  : 


32.  (a  +  Va  +  ^)(a  -  Va  -t-  b). 

33.  (Va  —  b  —  ^a){-y/a  —  b  +  Va). 

34.  (V2x  -  3  -  V3^)(V2x  -  3  +  Vs^). 

Express  as  radicals  of  same  order : 

35.  V2  and  V^3.  37.   Va^  and  VS. 

36.  VS  and  V^.  38.  V^  and  VS. 
Multiply  the  following : 

39.  V3,  -v^.  43.   ^,  V2.  47.   V^,  v^. 

40.  VS,  VS.  44.   Vl2,  x4.  48.   xl^ ,  ^P  • 

41.  V^,  Vs.  45.   Vc,  VS.  49.   ^y27^  V3^. 

.    42.   VS^,  VS.  46.   VS^,  VS^.  50.   VST^;  VS+^. 

73.  Division  of  radicals.  Direct  division  of  radicals  co- 
efficient by  coefficient  and  radicand  by  radicand  is  often 
possible. 

Thus  6V5-^3V3  =  2V|  =  §V15, 

and  3  Vox  -t-  2  VS  =  |  Va. 

Direct  division  of  radicals  when  the  divisor  is  a  radical 
expression  with  more  than  one  term  is  usually  very  diffi- 
cult. In  sucji  cases  a  rationalizing  factor  of  the  denomi- 
nator is  used.  We  then  carry  out  the  operation  of  division 
indirectly  by  resorting  to  multiplication. 

74.  Rationalizing  factor.  One  radical  expression  is  a 
rationalizing  factor  for  another  if  the  product  of  the  two 
is  rational. 


120  SECOND  COURSE  IN  ALGEBRA 

A  rationalizing  factor  for  Vz  is  Vt,  since  Vz  •  V?  =  7.  For 
"v/5  a  rationalizing  factor  is  ■v^25,  since  "Vb  •  ■v^25  =  5.  Similarly, 
Vs  —  V3  is  a  rationalizing  factor  of  V5  +  Vs,  as  their  product, 
(Vs  -  V3)  (V5  +  V3),  is  equal  to  5  -  3,  or  2. 

In  like  manner  (3  Vt  +  2  Vs)  (3  V?  -  2  V5  )  =  63  -  20  =  43. 

Two  important  radical  expressions  are  V^+Vi  and 
Va  —  Vb.  Two  such  binomials  are  called  conjugate  radicals, 
and  either  is  a  rationalizing  factor  for  the  other. 

Rationalizing  factors  are  used  in  division  of  radicals  as 
follows : 

,     ,        /;:        /r      V6V5      V30 
Example  1.    V 6  -^  V5  =  \_    ,_  =  -—-  • 

V5V5         5 

Example  2.   (6  V2  -  15  Vs)  .  3  Vs  =  ^^^"l^^j'^ 

3V5.V5 

_2VlO-5\^ 
5 

Examples.  (Vs +^)^(V5- Vii)  =  M±^^1M±2^ 

(V5-V3)(V5  +  V3) 

_  Vi5  +  ViQ  +  3  +  Ve 

5-3 

=  K^^  +  ^^  +  3  +  Ve)- 

Therefore  when  direct  division  of  radicals  is  impossible 
use  the 

Rule.  Write  the  dividend  over  the  divisor  in  the  form  of  a 
fraction.  Then  multiply  the  numerator  and  denominator  of 
the  fraction  by  a  rationalizing  factor  for  the  denominator  and 
simplify  the  resulting  fraction. 

This  rule  applies  in  all  cases,  while  the  rule  for  direct  division 
fails  when  dividing  a  real  radical  by  an  imaginary  number. 


RADICALS 


121 


EXERCISES 

Find  a  simple  rationalizing  factor  for 


1.  Vt. 

2.  3  Vs. 

3.  Vs. 

4.  Vi. 

5.  Vs. 


6.   Vs. 

7.  Vs-Vr. 

8.  V3  -  2.  . 

9.  3V6-2VII. 
10.   V3a-V2x. 

Perform  the  indicated  division  : 
16.   V8^V2. 

17.  eVio^Vs. 

18.  Vl2^V3. 

19.  V8--V24. 

20.  Va^H-Va^. 

21.  V2a^-v-V3a^. 


11.  Vx-3-V3. 

12.  Vic  -  3  -  2  V3^. 


13.  Va  —  ^  +  Va  -}-  6 

14.  V8+V2-V5. 

15.  V3H-Vi8-V^. 


22.  (Vi4-Vi0)--V2. 

23.  (2yiO-3V5)--2V5. 

24.  ( Va^  —  Va^) -^ VS. 

25.  8 --4  Vs. 

26.  8 --2  Vs. 

27.  24--3V3. 


28.   V3^V2. 

„  ,  ^.        V8      V3.\/4      V3.^4      ^27  •  16 

^^^"'""  vi = v^:^  ^  -^  ^  -^— 

29.   V5-V3.  35.  Vi-VJ. 


30.  V^--V2. 

31.  V8--V2. 

32.  Va-f-Va. 

33.  V2^^ 


4  -^  V  2 

O.  -f-  C  ViC. 


34.   V32-V2. 

R£ 


37.  V3^(V3-2). 

38.  V5--(V5+V2). 

39.  (2V3+V5)--(V3-V5) 

40.  (V7+V3)-(V7-V2). 


122  SECOND  CGHTESE  m  ALGEBRA 

Change  to  respectively  oqiaiYail^nt  fractions  having  rational 
denominators:  .,       ,  ,       , 

V5---^        ^.    2V5+"3V7  .    ,„    Va!-3+V3 
■'42:  —^  ■%.  y^.  48.        ^^^^     ■ 

V5;-V2,  Vcc+Vc  V2-V2 

Perform  the  indicated  division : 

50.  (  ViO  -  V5)-(  VlO  ^i'W):^^^  i>i^toiL/ii  61LI  nnol-i..<! 

51.  (i;'^^)-^<^-^-S^^).^'S  .SV-8V  .01 
53&'^( VW  4-  c  -  ^^)  V (:^i/a^c  +  V^).  .0 V --  oIV  l)  ^.TI 
53,n(^-^3,H7V2),^y3^  V2).               .FiV-^  siV  ,81 

54.  (  V5  +  V7^V2l3(  V|  -  V7).  .  ,v_  aV   .ei 

55.  Is  there  any  distinction  between  the  meaning  of  the 
direction  before  Eiercise  il^lid  of  that  bef of e  Exercise  5<)  ? 

56.  Does  3  -Vz  satisfy  a:?&-  6  a;  +  2  =  0  ?        on  ^jV  .iSi 

57.  Does  '^"^        satisfy  2x'-75x  +  161  ^  ?8S 

58.  Does  ^(B'  ±  Vto^)*satisfy ^3 i^^jc^^ ^  =  0  ? 

75.  Square  roQt  of  surd  expressions.  The  square  of  a 
binomial  is  usually  a 'trinomial.  However,  the  result"* of 
squaring  a  binomial  of  the  form  Va  +  V^  is  a  binomial  if 
a  and  h  ara  rational  number^.    Thus  ..^      /'      \y 

In  10-^2Y21^  lO^i^vtl^a  sgp  of  7  and  3,  and  ^1  is^the 
product  of  7  and  3.  Tnese  relations  anii  the  lact  tnat 
thfe^ooeffi^eHiof/the'^kdieftl  V2T  is  2  eiiake  Ue^to-ffiid 


A>i<i.ivM.  RADICALS    azooa;^  i2$ 

the  square  root  of  many  expressions  of  the  fomTa  ±V6 
by  writing  each  in  the  form  of  x±^^xy-\-y  and  thea 
extracting  the  square  root  of  the  trinomial  square  as  follows  f 

Example.   Extract  tne  square  foot  of  9  —  yoo. 

,  ^  __>/,,!,  y>',miLy6'iq  iloiiiw  anoift89iqz9  \imm 

Solution.   9-V66=J,^-^^A(^^  dfl-gjiodi  od  won  v/ifn  smiiq 
We  must  now  find  two  numbers  whose  sum  is  9   and  w;hose 

product  is  14.    These  are  7  and  2.'  '  -  "  ^  ~  "'  '"  rwiiy 

Then  9  -  2  Vii  =  7  -  i^^' ^f'^^^ -^ .  ^'^^ 

Heiice  th^  sqtmi^  rdot^^f  ^i^^^^^i^  "tg  (H^H)^.  ^^^^^  «I 
■    ■     ••   -iL^-j.Jjjn  iO    --   •   '    iijrl)  hoojaiohnn  yIibsIo 
.8ioio.Bl  oipXERCiSES-t  Qflj  fii  ginsioifiooo  odd 
ei  Find  the  positive  squafe  roots  in  Exereise^^d[*-:l^iri^'89'i  oT 

^'\B^k  V^-        ^.11-3  Vs.  ,; '      ^3  tMte   /n-^Lim 
4.  7-V40.  8.  Il-Vi20.  4  .Soj^fd 

12.  2  cc  +  2  Vx2  -  49.  14.  V^  +  3  VS  =  V?  4-^. 


13.  a  +  Va^  -  1.  15.'  V15  -  5  VS  =  ? 

'*         16.   V«  +  V^^^^^^^  =  ? 

17.   V  ??i"^  +  7?^  H-  2,  ?i  -f-  2  ??i  V m  +  2  71  =  ? 

Note.  In  the  writings  of  one  of  the  later  Hindu  mathematicians 
(about  A. D.  1150-)'we  find  a  method  of  extracting  the  square  root  of 
suTd.s  which  is  poetically  the.ipame  as  that  given  in^the  te?:t.§In 
fact,  the  formula  for  the  operation  is  given,  apart  froin  the  modern 

symbols,  as  follows :  Va  +  Vh  =  V «  +  Z>  +  2  Vah.  The  study  of  ex- 
pressitdTS  of  the  t^i^"  V  Va  i  ^6  had  been  carried  to  a  i^tto^tTeii^rk- 
able  degree  of  acQuracy  by  the  Greek,  Euclid.  His  researches  on  this 
subject,  if  original  with  him,  place  him  among  the  keenest  mathe- 
maticians of  all  time,  but  his  work  and  all  of  his  results  are  ex- 
pressed in  geometrical  language  which  is  very  far  reindved  f^oirPthe 
algebraijc  aymboli^iXJi,of  tp-d^  0  _  g  _  -  ^^  ^  .01 


124  SECOND  COURSE  IN  ALGEBRA 

.  76.  Factors  involving  radicals.  In  the  chapter  on  Factor- 
ing it  was  definitely  stated  that  factors  involving  radicals 
would  not  then  be  considered.  This  limitation  on  the 
character  of  a  factor  is  no  longer  necessary.  Consequently 
many  expressions  which  previously  have  been  regarded  as 
prime  may  now  be  thought  of  as  factorable. 

Thus  3  x2  -  1  =  (a:  V3  +  l)(a:  VS  - 1), 

and  4.x^-5  =  (2x  +  V6)(2x-^^'). 

In  this  extension  of  our  notion  of  a  factor  it  must  be 
clearly  understood  that  the  use  of  radicals  is  limited  to 
the  coefficients  in  the  terms  of  the  factors. 

To  restrict  the  use  of  radicals  in  the  way  just  indicated  is 
necessary  for  the  sake  of  definiteness.  Otherwise  it  would 
be  impossible  to  obey  a  direction  to  factor  even  so  simple 
an  expression  as  x^—y"^;  for  if  the  unknown  is  allowed 
under  a  radical  sign  in  a  factor,  x^  —  y^  has  countless 
factors. 

Thus     x^  —  y^  =  (x  ■\-  y^(x  -^  y) 

=  (x  +  y)  (Vx  +  Vy)  ( V.r  —  Vy) 

=  (x  +  y)  (Vx  +  Vy)  (Vx  +  V^)  (Vx  —  Vy), 

and  so  on  indefinitely. 

EXERCISES 
Factor : 

1.  ar^-11.  3.  a;«  +  3.  5.3x^-27, 

2.  3a^2-8.  4.0^^-12.  6.  6x«4-125. 
Find  the  algebraic  sum  of  the  following : 

2-Vb  2  x  +  c  x^  -^c^ 

'  a-b       V«-f  V^^*  *    V^-Vc        ^-c 

Solve  by  factoring,  and  check  the  results : 
9.  x'-b^Q.  11.  (T*  +  144  =  26x1 

10.  2x2-3  =  0.  12.  4.x'-{-c  =  x^+A:Cx\ 


RADICALS  125 

PROBLEMS 

(Obtain  answers  in  simplest  radical  form.) 

1.  The  side  of  an  equilateral  triangle  is  8.  Find  the  altitude. 

2.  The  side  of  an  equilateral  triangle  is  s.  Find  the  altitude 
and  the  area. 

3.  The  altitude  of  an  equilateral  triangle  is  24.  Find  one 
side  and  the  area. 

4.  Find  the  side  of  an  equilateral  triangle  whose  altitude  is  a. 

5.  Find  the  altitude  on  the  longest  side  of  the  triangle 
whose  sides  are  11,  13,  and  20.    Find  the  area  of  the  triangle. 

Hint.  Let  the  altitude  on  the  side  20  be  x,  and  the  two  parts  into 
which  the  altitude  divides  side  20  be  y  and  20—  y ;  then  set  up  two 
equations  involving  x  and  y,  and  solve. 

6.  Find  the  altitude  on  the  longest  side  of  the  triangle 

whose  sides  are  10,  12,  and  16. 

'      '  E n 

Fact  froTYi  Geometry.    A  regu-  /  \ 

lar  hexagon  may  be  divided  into  /  \ 

six  equal  equilateral  triangles  by  /  \ 

lines  from  its  center  to  the  vertices.     fI  2  \fy 

In  the  adjacent  regular  hexa-         \ 
gon,AB  =  BC  =  CD,  etc.   0  is  the  \        7 

center  and   OK  perpendicular  to  \/ 

^^  is  the  apothem  of  the  hexagon.  A  '     K       B 

7.  Find  the  apothem  and  the  area  of  a  regular  hexagon 
(a)  whose  side  is  18;    (h)  whose  side  is  s. 

8.  Find  the  side  and  the  area  of  a  regular  hexagon 
(a)  whose  apothem  is  30;    (h)  whose  apothem  is  h. 

Facts  from  Geometry.    The  volume  of  a  pyramid  or  cone  is 

—  >  where  a  is  the  altitude  and  h  is  the  area  of  the  base. 

The  altitudes  of  an  equilateral  triangle  intersect  at  a  point 
which  divides  each  altitude  into  two  parts  whose  ratio  is  2  to  1. 


126  SECOND  Ce^ESE'  IN  ALGEBRA 

9.  The  base  of  a  pyramid  is  a  square,  each  side  of  which 
is  10  feet.  The  other  four  edges  are  each  20  feet.  Find  the 
altitude  and  the  volume  of  the  pyramid. 

,3 [ , jjp^  r  jTJ;ie:  side  pf  an  equilateral  triangle  ift  ^^ , j|!in^  the  two 
parts  into  which  each  altitude  is  divided  by  the  othey  altitudes. 

'>i 'A'  regular-  tetrahedron   is '^^i^P'^  'ifJ^  to  oLiJ^jjUu  siiT  .£ 
pyramid  whose  four  faces  are  .Bs^kaif.^  hua  mFjih 

eq^ual  equilateral  trianglesJ  •rr.^tiiliirpo  iir>  io  oMh  dkM)[ii  . 
sl-The    altitude    of  a  regulajrf.j  no  obufi/B  }\\\^1   .a 
t%%ahedron , (>Z>^  in  the   adja-,os;  bar,  ,^/,tl  ojias^H  sao/lw 
§^^t  iigui-e)  ^eets,the,base,^t  ,,i,  ,,0  9/jijiB  ..iJ.li  .\rH 
<fc^e  point  where  the.-  altitudes  os  ahia/jLivib  Qbiijjjifi  j)j\[:)h[v/ 
of  the  base  intersect.  ..»7[oa  brf^\y  bits  tftgriiJloini  8/ioiVxri>o 

rsihedT^.    If  each, edge  is  24,-    =''''^N<^'   ''■'<\^^)}^^''^>:yC'ff 
find  C/!^'VK'^a,ndt;-^  lastly,  tlm^^yi  A    av^''^J?*v^'^^<>'\V.  "^^'^ 
altituae  Z)^.  oJiii  bobivib.;:Jt:l   v(^i;iii>8<{AaKyiI  ud 

J%.  Find  the  altitudeandvai  ;:''^""":'  '^"''""f"  *" ^" . ' ,' 
ui^e  of  a  regdlar  tetrahedron  each  of  whose  ed^esis  ^6  iiiclies. 

ik  Show  that  the /ititude  and  the  voluthe^of  a»tfegulartetra^ 

hedron\j^hose-^4^is  e  are  respectively  %  "V^  and  tt:  ^(2. 

MISCELLANEOUS  EXERCISES 

f  r*  1  iRediice!  fc6  respectively  equivalent  fractions  haviingf  tatibnal 
denominaltoaB  i^i'MiJoqij  o^.uii  ^/   (A),  -(jii  rii  tiirj(l,jo<j.B  abOiiYV  (^^  i 


Find  the  positive  square  root"  of  the  following : 

7.  28  +  10V3...^ li_i2.a.-5I-=.  l^yiS. 

8.  30  +  12  Ve.  J~V  10.  c^  +  c  +  2  c^  V^. 

11.  Show  that  X  —  ..  .■ ,  ^  .    .;  isa  root.^^ii^^  —  7  a?  +  1  =  0. 

12.  Show  that  X  =        ^^„       :^^rB  ruuls  o^f^x"^  +  6  a;  =  1. 

o  ■ 

13.  Show  that  x  =  3  a  ±  2  Vc  are  roots  of  a;^  —  6  ax  +  9  a^ 

1.     Qv.        .T.  .  -  ^^>  +  V60,^^  +  a^^>^  .  ,     .  ,, 

14.  Show  that  x  =  ig^^  root  of  the 

equation  3  ax^  +  abx  =  5  b.         ~  ~ 
Simplify  : 

15.  2V18H-3V5- V8-*V32.  17,  V3  -  V6. 

— : , — -.  ^  ., .lo 

16.  sVf +  2\/^-^Vf  H-'^^^^-  18-   V5H-3V2. 

20.  (5  V7  -  ^)i;^(v^  +  V7)r  ^^ 

21.  sVi  +  Ss/S+f  V96-V66|. 

22.  ;     ^^       -  ^,        ^  ^^  <,-) 

25.  03  .  i  (a^  -  xYH-  2  ^)  -  («'  -  ^')^. 


26. 


(x^)^ 


his) 


x'(S  x^)  -  (x^  4-  5)  4  x^      x^  +  20 
(x*)2  •        x^ 


128  SECOND  COURSE  IN  ALGEBRA 


(x^  -  If 


28. 


29. 


30. 

X 


x^ 

x\ax^- 

x'-l 

-\-l)2x 

{xy 

x^^'nx^- 

x' 
-1  -  x''  . 

2ax^^-^ 

(x^y  . 

,2  a 


31. 


32. 


x-'(-  2x-^)-(x^-\-S)(-  5x-*) 

(x-y 


2  e"'^  +  1 


33.    ^^ ^pr-^ 

■\x 

•^  +  1 


CHAPTER  IX 

FUNCTIONS  AND  THEIR  GRAPHS 

77.  Functions.  One  of  the  most  important  concepts  of 
mathematics  is  the  notion  of  function. 

As  the  term  is  used  in  mathematics  the  basic  idea  is  the 
dependence  of  one  quantity  upon  another  or  upon  several 
others.  Countless  functional  relations  exist  in  everyday- 
affairs.  The  velocity  of  a  falling  body  is  a  function  of 
the  time  since  it  started  to  fall,  the  interest  on  a  definite 
sum  of  money  is  a  function  of  the  time  and  the  rate,  and 
the  quantity  of  water  transported  yearly  by  the  Mississippi 
is  a  function  of  the  rainfall  (and  the  snowfall)  in  its  basin. 
Relations  such  as  these  can  often  be  expressed  either  exactly 
or  approximately  by  equations.  For  a  body  falling  from 
rest  near  the  earth's  surface,  s  =  16  t\  Here  the  distance  in 
feet,  s,  is  expressed  as  a  function  of  the  time,  ^,  in  seconds. 
In  y  =  7?  -\-  2>  X  -\-l  the  value  of  y  is  a  function  of  x.  Such 
a  relation  is  often  represented  more  clearly  and  strikingly 
by  a  graph  of  the  equation  than  by  the  equation  itself. 

78.  Names  of  functions.  A  function  is  called  linear,  quad- 
ratic, or  cubic  according  as  its  degree  with  respect  to  the 
unknown  or  unknowns  is  first,  second,  or  thirS  respectively. 

Thus  4  a:  —  7  is  a  linear  function  of  a: ;  3  x^  —  8  a;  +  1  is  a  quadratic 
function  of  x ;  and  a:^  —  3  a:^  +  a:  —  10  is  a  cubic  function  of  x. 

In  the  study  of  functions  the  unknown  is  often  called  the 
variable,  since  from  this  point  of  view  the  problem  is  not  so  much 
the  finding  of  an  unknown  as  it  is  the  study  of  the  changes  of  a 
variable  quantity. 

129 


130 


SECOND  COUESE  IN  ALGEBRA 


79.  Notations  for  a  function.  After  a  function  of  any 
variable  x  has  once  been  given  it  is  usual  to  refer  to  it 
later  in  the  same  discussion  by  the  symbol  /(.r),  which  is 
read  the  function  of  Jr,  or,  more  briefly,  /  of  x. 

80.  Linear  functions^  THd  dxpresfeion  3  a;  4-  2  is  a  func- 
tion of  x^  and  the  value  of  this  binomial  varies  with  x. 
The  following  table  gives  a  partial  view  of  the  relative 
^change  of  val,u^s,,J?ptw^l^|}5  a::,,an4,  tl;^^  fung|4^^7§^ 


fioij.mtr^    t 


If 


n,i. 


i? 


theii/(i)  =  8x  +  2=i^l0!  tt^  v4i4i  im^'  -iQ'!   i'5  •   'i8-'Iiili(j)I 


,•>., 


.?, 


u. 


This  relation  can,  be  xepo^^^anted  graphically  l;>yiTisuigtih,e 
same  ir-axis   as  beforQj 
(section  44)  and  using 
the.j/-axis  as  the  func- 
tion axis;  that  is,  laying  ; 
off  values  of  x  horizour.  , 
tally  aad  corresponding  .j 
values  of  the  function^  \ 
^x  +  2  vertically.  Thev 
graph   resulting    from,  ^ 
the  above  t^J^le  iqf  y^U^k 
ues   is    shown  in    the, 
^CQmpanying     figu^p,,,. 
J^ic^  bja  show^  th^^t,  M 
the  graph  ofr  a  Imea^ , , , 
function    is  ,  always  „  a 

straight   Un^-,,j.,„^,^  ^l^^^f,y  ^>  ^j  or       .    i   •  ,  j;  -  •'■•,  I.jik  ;x  lu  aolUnn^ 
Mift    i)'>rii;')    n'ttlii    «f    iiwoii /({EXERCISES 'i'fojiult    ^o    \Imi\<'.    'mH    nl 

il'UMU  OK  Jtur  yi   n!''iifo'iif  -Mit  /■'•'i      ■  •   ',"■""!  HfiM,  rnn-fj  -yiiiiij  ^'tUlnhivf 

1.  Construct , the  graph  of  the  function  3  a:  ~L.,    ..     ,, 

2.  Construct  the  graph  of  the  function/^ w,rln .3.  .Mjiirijv 


ru3srcTiONS  axd  their  graphs 


131 


81.  Quadratic  functions.     The  function  x^-{-x—6  may 
be  represented. graphically  by  proceeding  a^  :^ollows: 


Qd! 


a 


1;h^ii'J^(i9±i:?ar*'{P'dt-t^6t=it.q(i6f  \>i^  :  u.4 


'i'iM    k,M.[l';!l! 


^S 


rhrrrrr 


a^-l'. 


^e 


«' 


-6 


-4 


^ 


s 


■6!' 


-^f' Plotting  the\ 
*^'6ints      corre- 
sponding to  the 
numbers  in  thj^^'r 

■-'..: 

■ 

T-rr 

•In 

IT, 

"\ 

I  ,  1  1  j 

up-: 

;t8y 

I<Jif  ir;    \ 

■jM'/ 

•  ^ ;  i  ,' 

•   ', 

!     '> 

if 

i       , 

GRAPH 

OF 

ii 

-lo 

iiq: 

1   / 

table  we  obtain  ^- 

i'l'J 

i'l'J 

\i/ 

m- 

^^f. 

;aj- 

B  ; 

tn  ' 

;  ,( if 

•     r  i 

/' 

the  accompany-'  - 

-  ""  ■\. 

jil 

A 

::■: 

:n 

' 

.1 
0 

!!''! 

"  li 

/  ' 

ing  graph*  ^-^oUoi 

8S 

•o,.i 

\ 

'  r  1 

I  '/ 

(i   V 

!];;■ 

j~-Th;e  -  g 
df  a  quae 

ttmvtK 

1 

1  tip  11 

.ratio 

r 

I,  _ 

V- 

- 1-^ 

/ 

i 

\ 

/ 

fmnctionii 

ijone  1 

\ 

/ 

. .  _ .  1 

'• 

\ 

/ 

.  1 

y' 

variable     is     a 

3 

c 

. 

^-' 

J    -i    0 

i      /2       3 

,,:■■ 

[fiurve  called  K  . 

\ 

/ 

liurabola.  Itmayl 

0  /   ;.> 

\ 

~ 

)  1  ' 

~^ 

/ 

:ib 

l!<.>' 

f'.' ' . 

•■■■'- 

■>■;  - 

'    ■        ;    '» 

\ 

b^  sha^ 

r  ierj 

--, 

\ 

\y 

'flatter  thai 

a  the; 

hV 

; 

^^ 

.■ : :  ! 

■]'■ 

i  '  ' 

' "  j 

accompanyijig^ 

1\ 

-h 

'-J 

!!) 

p* 

' 

graphs  jbd 

+    ^  ' 

i  ' 

^n..^ 

'  .1 

V'-i- 

,- 

^  y.L-  :     I    \i 

the  same  general  shapfi^anfl-the  opening  may  be  upward 

or  downwfi^d^  .J'.i:...ii     i  ^no    :ji[}     nl 

f^-j   T    j  EXERCISES   »i^^    8088010    OYlir'o    OflT 

^1    ri^^T  L     „,      «  .:..  ^  •„      .:;^!ffiii      6^-nI}       aixB--^; 
Uonstruct  tna  graph  of  the  lollowmg :      ^  r 


7.  A  body  falling  from  rest  near  the  earth's  surface  obeys 
the  law  s  =  16t^,  where  s  is  in  feet  and  ^  is  in  seconds.  Con- 
stwt(&t  the  gi?^h  4ii  fit)  =  16 1^.  ii  ^     .xioirniiebxii    alliil   Ji    1 . 


132 


SECOND  COUKSE  IN  ALGEBRA 


-  Note.  In  the  study  of  analytical  geometry  one  takes  up  system- 
atically the  curves  which  represent  equations  of  the  various  degrees 
beginning  with  the  simplest.  It  turns  out,  as  we  have  already  seen, 
that  the  linear  equation  is  represented  by  a  straight  line..  Equations 
of  the  second  degree  in  x  and  y  lead  to  the  so-called  "conic  section." 
One  of  the  most  interesting  and  important  aspects  of  the  graphi- 
cal method  is  the  fact  that  the  simplest  equations  correspond  to  the 
most  useful  curves  both  in  pure  science  and  in  nature.  The  com- 
monest curves  in  nature  are  the  circle  and  the  parabola.  Their 
equations  are  the.  very  simplest  equations  of  the  second  degree. 

82.  Graph  of  a  cubic'  function.  The  graph  of  a  cubic 
function  is  obtained  in  the  same  general  way  as  that  of 
a  quadratic  function.  The  function  a^—  5x+^  may  be 
represented  graphically  by  proceeding  as  follows: 


If                                     x  = 

-    4 

-3 

-2 

-1 

0 

1 

2 

n 

3 

then/(a;)=a:«-5a:  +  3  = 

-41 

-9 

5 

7 

3 

-1 

1 

6i 

15 

Plotting  the  points 
corresponding  to  the 
numbers  in  the  table 
(except  the  first  and 
last),  we  obtain  the 
points  ^(-3,  -9),  B, 
C,  D,  E,  G,  and  II, 
in  the  order  named. 
The  curve  crosses  the 
a:-axis  three  times : 
once  between  1  and 
2,  again  between  0. 
and  1,  and  a  third 
time  between  —  2  and 
—  3.  Above  If  the  curve  rises  indefinitely,  and  below 
A   it  falls   indefinitely.     In   each   case   it   becomes   more 


F 

GR 

WoF 

1 

f(3 

f)  = 

x'- 

5£C 

+3 

/ 

P 

/ 

/ 

fN. 

Ih- 

Yn 

/ 

/ 

/ 

\ 

/ 

\ 

^ 

X 

X 

f 

^  V 

I 

i 

-1 

6 

\ 

\ 

y 

1 

\ 

E 

1 

1 

1 

1 

1 

f 

o 

\ 

A 

F' 

FUNCTIONS  AND  THEIR  GRAPHS  133 

and  more   nearly  straight  as  it  recedes  from  the  a:-axis, 
never  crossing  either  axis  again. 

In  forming  a  table  of  values  two  pairs  are  sufficient  for  a  linear 
function,  but  more  are  needed  for  a  quadratic  function  and  still 
more  for  a  cubic  function.  Usually  the  higher  the  degree  of  the 
function,  the  more  points  are  needed  in  constructing  its  graph.  It 
should  be  noted  that  in  making  a  good  graph  the  number  of  points 
is  not  so  important  as  is  their  distribution,  which  should  be  such 
as  faithfully  to  outline  the  entire  curve.  Where  the  graph  curves 
rapidly  or  makes  sharp  turns  the  points  should  be  close  together. 
Such  places  are  difficult  to  locate  before  the  graph  is  constructed ; 
hence  one  should  make  a  table  of  values  which  appears  to  be  suffi- 
cient and  plot  them.  Then  inspection  of  the  plotted  points  will 
usually  show  where  sharp  turns  or  rapid  curvature  exists.  The 
table  of  values  should  then  be  properly  extended  and  the  additional 
points  located.  Repetition  of  this  last  step  will  enable  one  to  draw 
a  graph  which  accurately  pictures  the  variation  of  the  function. 

It  should  be  observed  here  that  scales  on  the  two  axes  need  not  be 
the  same.  Some  experience  is  required  to  choose  for  the  two  axes 
the  scales  which  are  best  suited  to  bring  out  clearly  the  shape  of  the 
curve.  In  general  the  graph  should  be  drawn  to  as  large  a  scale,  in 
both  directions,  as  the  size  of  the  paper  permits.  What  that  will  be 
for  each  axis  can  be  decided  by  inspecting  the  table  of  values.  For 
example,  when  the  dimensions  of  the  preceding  graph  are  once  deter- 
mined one  can  see  from  the  table  that  all  values  of  x  are  easily  repre- 
sented but  that  it  is  undesirable  to  try  t.o  represent  the  values  of  the 
function,  —  41  and  15. 

EXERCISES 

Construct  the  graph  of  the  following : 

1.  x^-Sx-^l.  3.  x»-4cc-2. 

2.  x^-Sx  +  2.  4.  £c*-ll£c2  +  24. 

Note.  The  notion  of  a  function  is  one  of  the  three  or  four  most 
fundamental  ideas  in  modem  mathematics.  Only  the  simplest  ex- 
amples are  given  in  this  book,  but  many  others  involving  expressions 
of  the  utmost  complexity  have  been  studied  by  mathematicians  for 
many  years.   An  important  reason  for  the-study  of  functions  is  found 


134  SECOND  COURSE  IN  ALGEBRA 

i|Q,  the  fact  that  all  kinds  of  facts  and  principles  which  we  meet  in  the 
study  of  nature  can  be  expressed  symbolically,  by  means  of ,  functions, 
and  the  discovery  of  the  properties  of  such  functions  helps  us  to 
understand  the  meaning  of  the  facts.  A  cohipl6te  understaiidiiig  of 
the  laws  of  falling  bodies,  light,  electriditj^,  o'l* 'teoiind ' could  tLevei^%(e 
reached  Without  the  study of  <te  '  Blattfeiii^W^k-'fi^'^ 
these  phenomena  suggest.  ^^  bsij^irr  str  p.:imo(i  sioiu  eiiJ  .noitoiiii't 

When  the  electrician,  the  architect,  or  the  artilleri'st  tnefefts  a 
problem,  he  frequently  must  represent  quantities  by  letters.  The  x 
and  the  y  may  represent  the  measures  of  objects  in  nature,  but  the^ 
solution  has  become  merely  an  operation  of  algebra.  As  students 
of  algebra  we  are  not  concerned  with  the  origin  of  the  functdon  or 
expression,  but  merely  with  the  iiumerical  determinatloitucrf  -eonael 
unknown  or  the  simplification  of  some  expression.     '.Iq  hnn  in-^ro 

^_        _  ,   .       -  ,        .       ■:      i.-       •■:    'VJ     (flRl!^-;     ')'i-)ii^//    7/Oii«     {lli\lLi>.i( 

83.  Graphical  solution  of  equations  in  one, unknown.  Jn 

of  the  preceding  sections  is  their  use  in  solving  equations^ 
in  one  unknown.  The  ideas  involved  can  be  made  clear 
by  questions?  <bii  -the  graphs  of-  sections  ^  80, '  81,  and  ^82. 

'>il:l  10 '<i[iidr.  od^  \hsif)h  ]'■:       .  '  ■    ifjiri  7«')d  fyw  ii;uiiw  fcj>i^Or'-  e»ni 

m  ^ofno?.  r>  ^v;h>\  ?,o  o*  a^OItAL  EXERCISfifi^'' ^'  ^'^*  iBiJ-^no^  xiT    .svtjjo 

ndJIi;i,.ij;n:t  JxMiV/'    -^iuniKf  v,(ns'f  ^^.  i   k,  '•^■l^'niitJ)  .^.^\vyV^vuV   AaoA 

X.  ¥rom  the  graph  m  sectibn  80  id^termine,the  value fOi  tna 

function  ^  3  x  -f-  2  at  the  'ppmt  yrhev^e  it^  ,g;^'f!'P,^  iC?^^^s^^.^  i^h^  ,«?7^fi9f> 

3.  Does  this  value  of  aa  satisfy  the  equation  3  a;-}- 2. *i)0t2n! 
8.  Solve  3a5  +  2  =  0  without  reference  tb  th^' gtaphl'^  f''^"»« 

4.  What  point  on  the.  gi;^ph  represents"  the  root  of 
3a; +  2  =  0?  ^'^^^'' 

5.  Irom  the  graph  in  section  81  determine  the  vulues  ui 
the  function  x^  ^  x  —  6  at  tHe  points  where  its  graph  crosWs 
the  a;-axis.    .*^-  4-  '^  -  +  ''-^  "  '  '    -^ 

6.  Do  these  values  of  «  satisfy  the  equation'^4^'5'— 6ii/0? 

7.  Solve  x^-{- x—6  =="(y  without  referving'to' tie  graphl '.  ^  ^ 
,  ,i8,„TO^,,piQip^,pp,,tia^  ngxaph   repre^Qftii  tbft  .^ootBiipf, 


F^KCTIONS  .iLasrE>  THEIR  GRAPHS  135 

9.  From  >tbe  giapiiiin  action  82  determine  tlie  values  of 
the  functioixiqff!^  5.;?^  rb  3  at  the  points  where  its  graph  crosses 

It).  Do  these  values  of  oj  mak^-the  functioa  i^-^&As^f  B 

equal  zero  ?  .3ini  gi  eifia 

11.  What   method   could   beHsi&d'  to   solve   the  ^qiuatloii 

84.  The  process  of  graphical  solution.  From  what  pre- 
cedes, it  is  apparent  tlmt  tfee  steps  in.  the  graphical  solution 
of  an  equation  ip  one  unknown  are:  ,  m^ 

^,,  Transpose  the  terms  so  that  the  right  member 4$  ^^(%idBi}pe 

^iilGrmjpJp  ^HJkM^^9V(  in  the,  left  {m^mf^er,    ,   : , : ;    , ,  j  r      , ,,, . r... ; t 

The  values  of  xfor  the  points  wlipretfi^  gi^ph-  crp,^s&»,t^, 

X-axis  are  the  real  roof$  ^f  the  eqpation.\Q  imyi  dirjji8i!00  erij 

The  algebraic  solutions  of  a  linear  equation  and  a  quad- 
ratic equation  in  one  unknown  are  so  simple  that  except 
for  the  purpose  of  illustration  their  graphical  solution  is 
comparatively  unimportant.  The  algebraic  solution  of  cubic 
and  higher  equations,  except  in  simple  eases,  is  much  more 
difficult  than  the  solution  of  the  quadratic  and  is  never 
iresented\ij^s^an  elementary  course.  For  this  i»^asOil  and 
ijoT  the  insiglit  it  gives  into  equations  in  general,  the 
Graphical  solution  of  the  cubic  and  higher  equations  is 
i|mportant  Ti!q[d: - fflnminating.  For:  subfe' equations  if  the 
method  of  factoring  fails,  the  graphical  method  is  the  only 
method  open  to  the  student  at  this  point  in  bis.  progl^ss. 

I,  x^-Tx  +  S,  f^jO.xo'  a  =  't  8ort4.[^?  -8^  =  0.   ,  )(c.  -  5.) 


136 


SECOND  COURSE  IN  ALGEBRA 


85.  Imaginary  roots.  An  equation  of  the  second  or 
a  higher  degree  often  has  imaginary  roots.  Such  roots  can- 
not be  obtained  by  the  graphical  methods  so  far  considered. 
A  study  of  the  graphs  which  follow  will  make  clear  why 
this  is  true. 

Consider  the  following  equations: 

a^-4:x-5  =  0,  (1 ) 

2^_4:^;  +  4  =  0,  (2) 

^_42;  +  13  =  0.  (3) 

The  graphs  of  the  functions  in  the  left  members  of 
equations  (1),  (2),  and  (3)  are  given  in  the  accompanying 
figure.  The  three  functions  differ  only  in  their  constant 
terms,  for  9  added  to 
the  constant  term  of 
(1)  gives  the  constant 
term  of  (2),  and  9 
added  to  the  constant 
term  of  (2)  gives  the 
constant  term  of  (3). 
Apparently,  as  the 
constant  term  is  in- 
creased the  graph  rises 
without  change  of 
shape  and  without 
motion  to  the  left  or 
to  the  right. 

From  the  graph  the 
roots  of  a?  —  4  x  —  5  =  0  are  seen  to  be  5  and  —  1.  These 
results  are  obtained  from  factoring ;  a^  —  4a:  —  5  =  0,  or 
(^x  —  5)(a;  + 1)  =  0.   Whence  x=5  or  —  1. 

If  we  imagine  curve  (1)  to  move  upward,  the  two  roots 
change  in  value  and  become  the  single  root  of,  curve  (2), 


F 

V                                             ~! 

\ 

Vf-                                        t1 

Iv                     71 

\  i  \                           /  '  / 

\*                         '  '7 

3aS                    ttt 

Cv^             -  U 

SjuA.       9n                       Z-J-i- 

\5^v  6       -/^Q 

Xx^^.'X  yJ-l- 

yyj^i-^'^ztz 

\    "^  4                •''    / 

X                    \       N    (2)   ..'      /                    X 

-2  -\  0  1    2    3      A 

^  y4          1          y'                                                '     ■■■ 

-8*^4-^ 

F     1 

FUNCTIONS  AND  THEIR  GRAPHS 


137 


which  touches  the  a;-axis  at  a  point  where  x  equals  2.  Solv- 
ing 2j2  —  42^  +  4  =  0  by  factoring  gives  (x  —  2) (a;  —  2)  =  0. 
Whence  x=2. 

If  we  now  imagine  curve  (1)  to  move  still  farther  upward 
from  its  position  (2),  it  will  no  longer  cut  the  a^-axis. 
Further,  when  the  curve  reaches  the  position  of  (3)  it 
does  not  cut  the  a;-axis  at  all,  and  hence  cannot  show  the 
values  of  the  roots  of  the  equation  a:^  _  4^_j_13  _  0^  as, 
in  fact,  it  does  not.  The  graph  does  show,  however,  that 
the  value  of  a:^  —  4a;  +  13  at  the  lowest  point  of  the  curve 
is  9.  This  means  that  for  every  real  value  of  a?,  positive 
or  negative,  x^  —  4:x+l'^  is  never  less  than  9.  The  graph 
of  (3)  makes  clear  that  no  real 
number  if  substituted  for  x  will 
make  a:^  —  4  a;  + 13  equal  zero. 

It  can  be  shown  by  the  method 
of  section  87  that  the  roots  of 
a^-4ic  +  13  =  0  are  2  +  3V^ 
and  2-3^/^. 

Note.  It  required  the  genius  of  no 
less  a  man  than  Sir  Isaac  Newton  first 
to  observe  from,  the  graph  of  a  function 
that  two  of  its  roots  become  imaginary 
simultaneously.  He  also  saw  that  an 
equation  with  two  of  its  roots  equal  to 
each  other  is,  in  a  certain  sense,  the  lim- 
iting case  between  equations  in  which 
the  corresponding  roots  appear  as  two 
real  and  distinct  roots  and  those  in 
which  they  appear  as  imaginary  roots. 

86.  Imaginary  roots  for  a  cubic  equation.  If  we  attempt 
to  solve  2;^  —  2^;  —  4  =  0  graphically,  we  obtain  the  graph 
of  the  accompanying  figure.  The  curve  crosses  the  2;-axis  at 
x=2.   This  is  the  only  real  root  the  equation  has ;  the  other 

RE 


F 

0 

J 

] 

X 

' 

X 

_9 

0 

3    1 

/ 

/ 

/ 

^ 

1 

/ 

^ 

J 

/ 

-5 

/ 

GRA 

PH 

OF 

^ 

n 

xS= 

X'- 

-Zx 

^4 

"r 

_ 

F 

138  SECOND  COURSE  IN  ALGEBRA 

two  are  imaginary.  The  roots  can  here  be  obtained  by  fac- 
toring a^-2x-4:  =  0;  thus  (x-2)(x^'{-2x  +  2}  =  0. 
The  roots  of  2:24- 2  a:;— 2  =  0  are  the  two  imaginary  roots 
of  the  euhic  equation. 

EXERCISES 

As  far  as  possible  solve  graphically,  finding  results  to  one 
decimal  place : 


1. 

x^-5x-9  =  0. 

6.  x^-4:X-^5  =  0. 

2. 

x^  =  4.x-5. 

7.  X*  -  7  ic  +  4  =  0. 

3. 

a:^  —  3  X  +  4  =  0. 

8.  ic*-4x»  +  12  =  0. 

4. 

a;3  +  ic  -  4  =  0. 

9.  a;*  =  10x^-9. 

6. 

x^-lx'-2x  +  S  = 

=  0. 

10.  x^- 2x^  =  0. 

CHAPTER  X 

QUADRATIC  EQUATIONS 

87.  Solution  by  completing  the  square.  An  equation  of 
the  form  a7?  -{-  62;  -f-  c  =  0,  where  a,  5,  and  c  denote  num- 
bers or  known  literal  expressions,  is  called  a  quadratic  equa- 
tion. Any  such  equation  can  be  solved  by  the  method  of 
completing  the  square.  This  method  gets  its  name  from 
the  fact  that  in  the  course  of  the  solution  there  is  added  to 
each  member  of  the  equation  a  number  making  one  member 
a  perfect  square. 

ORAL  EXERCISES 

What  terms  should  be  added  in  order  to  make  the  following 
expressions  perfect  squares  ? 


1.  x2  +  2ic  +  ? 

5.  ir2+a^  +  ? 

9.  cc2  +  |x  +  ? 

2.  x^-2x^l 

6.  a;2_3^_^9 

10.  x^-\x+l 

3.  a;2_6^4-? 

7.  £c2_^fx  +  ? 

11.  cc^  +  |x  +  ? 

4.  ic2  +  8x  +  ? 

8.  a^2_4^_|.? 
EXAMPLE 

12.  x'-\-ax-\-'i 

Solve5ar^-3x 

—  2  =  0  and  check  the  result. 

Solution.                                     5  :r2  -  3  a:  -  2 

« 

Transposing,                                      5  a;^  —  3  a: 
Dividing  by  the  coefficient  of  a:^,       x^  —  \x 
Adding  (—  ^-^y  to  each  member, 

=  0.                         (1) 
=  2. 

or 

139 


140  SECOND  COURSE  IN  ALGEBRA 

Extracting  the  square  root  of  each  member, 


X 


h 


±/zT- 


Whence  ^  =  i%  ±  tV  =  1  ^^^  ~  §• 

Check.    Substituting  1  for  x  in  (1), 

5- 12 -3- 1-2  =  0, 
5-3-2  =  0, 
0  =  0. 
Substituting  —  f  for  x  in  (1), 

5(-f)^-3(-f)-2  =  0, 

1  +  1-2-0, 

0  =  0. 

The  method  of  solving  a  quadratic  equation  illustrated 
in  the  preceding  example' is  stated  in  the 

Rule,  Transpose  so  that  the  terms  containing  x  are  in  the 
first  member  and  those  which  do  not  contain  x  are  in  the 
second. 

Divide  each  member  of  the  equation  by  the  coefficient  of  x^ 
unless  that  coefficient  is  +1. 

In  the  equation  just  obtained^  add  to  each  member  the 
square  of  one  half  the  coefficient  of  jc,  thus  mahing  the  first 
member  a  perfect  trinomial  square. 

Rewrite  the  equation^  expressing  the  first  member  as  the 
square  of  a  binomial  and  the  second  member  in  its  simplest 
form. 

Extract  the  square  root  of  both  members  of  the  equation^ 
and  write  the  sign  ±  before  the  square  root  of  the  second 
member^  thus  obtaining  two  linear  equations. 

Solve  the  equation  in  which  the  second  member  is  taken 
with  the  sign  +,  and  then  solve  the  equation  in  which  the 
second  member  is  taken  with  the  sign  — .  The  results  are  the 
roots  of  the  quadratic. 


QUADEATIO  EQUATIONS  141 

Check.  Substitute  each  result  separately  in  place  of  x  in  the 
original  equation.  If  the  resulting  equations  are  not  obvious 
identities^  simplify  each  until  it  becomes  one, 

EXERCISES 

Solve  by  completing  the  square  and  check  real  results  as 
directed  by  the  teacher : 

1.  £c2  _^  4£c  -h  3  =  0.  9.  6ic2  +  ic  -  35  =  0. 

2.  0^-2^-8  =  0.  10.  12x2 -25a; +  12  =  0. 

3.  ^2  -  5  -  2  =  0.  11.  3  ^2  _^  8  ^  +  4  =  0. 

4.  x2-h2=-3a:.  12.  x-{-2  =  ^x\ 

5.  2x2  +  5^2  +  3  =  0.  13.  x-4  +  x2  =  6-2x2  +  8x 

6.  3x2  +  7x-6  =  0.  ,^^+14 

7.  2^2- 3x- 5  =  0.  4      ' 

8.  5x2-7a!-6  =  0.  15.  (3x  -  2)^ +  (x  -  1)^  =  1. 

In  Exercises  16-25  obtain  results  to  three  decimal  places : 

16.  £c2-12x  +  31  =  0. 

Hints.   By  applying  the  rule  we  get 

X  =  6  +  VS, 
and  X  =  6  —  Vs. 

,    From  the  table  on  page  274,   V'5  =  2.236. 
Hence  we  get  the  result,  to  three  decimals, 

a;  =  8.236  and  3.764. 


4~  3 


17. 

x^-x-^l^^. 

22. 

Sx' 

18. 

n^  +  2  =  5n. 

23. 

x'- 

19. 

3x2_l2x  +  9  =  2. 

Q 

20. 

5a;2  +  8x  +  2  =  0. 

24. 

21. 

10-2x'  =  x^-Tx. 

25. 

(y-\ 

25.  (2/+l)(2/  +  2)  =  32/(y-4). 


142  SECOND  COUESE  IN  ALGEBRA 

26.  £c*  -  3  cc^  +  2  =  0. 

Note.   This  is  not  a  quadratic  equation,  but  many  equations  of 
this  form  can  be  solved  by  the  methods  applicable  to  quadratics. 


i- 


Solution. 

x4  -  3  a:2  +  2  =  0. 

xi-Sx^=-2. 
x^-3a:2+|=-2  +  |  = 

x^-^=±h 
a:2  =  2  or  1. 

Whence 

x  =  ±V2,  ±1. 

Check  as 

usual. 

Note.  It  should  be  particularly  observed  that  the  equation  of 
Exercise  26  has  four  roots  instead  of  two.  In  general  an  equation 
has  a  number  of  roots  equal  to  its  degree.  Thus  the  equations  in 
Exercises  29  and  31  have  six  and  eight  roots  respectively,  although 
some  of  them  are  imaginary  and  the  student  at  present  should  not 
be  required  to  find  them  at  all. 

27.  ic*-5a;2  +  4  =  0.  32.  6  cc^  -  11  ic^ -j- 3  =  0. 

28.  x'  -  ISoc^  +  36  =  0.  ,  33.  4m«  -  15  =  7m«. 

29.  x^-j-S  =  9x\  34..  x^-4.x^-\-S  =  0. 

30.  x^  -  7x^  =  8.  35.  2/'  +  7/  =  8. 

31.  a;«- 17 0-^  +  16  =  0.  36.  m*-4  V2m2  +  7  =  0. 

37.  (x^  -  2xy-  7 (x''  -  2x)  =  -12. 
Solution.    Let  x^  —  2  x  =  y. 

Substituting  y  for  x^  —  2xy  we  obtain 

Solving,  .y  =  3  and  4. 

Then  a;^  -  2  a:  =  3. 

Whence  a:  =  3  and  —  1. 

Also  a;'*  —  2  .r  =  4. 

Whence  •  x  =  l±  V5. 


QUADRATIC  EQUATIONS  143 

In  Exercises  38-41  do  not  expand,  but  solve  as  in  Exercise  37 : 

38.  (x-iy-\-4.(x-l)=5. 

39.  (x^-4.xy~5(x'-4.x)-24:  =  0. 

40.  S(2f-^Syy-7(f  +  S2j)=20. 

41.  (x  -  ^y+  Jx-^-5  =  0. 

42.  aa^  +  6x  -f  c  =  0. 
Solution,      ax^  +  bx  +  c  =  0. 

x^  ■{ —  X  =  —  -' 
a  a 


.^b      ,  /  bV       J2       ^      j2_ 

a:2  +  -a;+    — -    =—---  =  — - 

a         \2  a/       4  a^      rt  4 


—  iac 


,     b  Vft^  —  4  ac 


2a  2 


___6_      \^b^  -  '^  ac  _  -  b  ±Vb^  -  4:  ac 
^~     2a'^         2  a         ~  2  a 

43.  x^-\-Sax-h2a^  =  0.  46.  *cc2-a(J+l)a?  +  ^t2=0. 

44.  2f  +  b2j=6b\  ,.      h  ^        b        , 

45.  bx^  +  x  =  l+bx.  2a  2a 

48.  cx^  -{(^-itd)x  +  cd  =  0. 

49.  a^ic^  —  aj  (a  +  ^)  Va^  +  aZ>  =  0. 

50.  3a.*-aV-2a^  =  0.  „    3   ,     ^  1    .     9a^ 

53.  7cc^  — 2aa:-h-a2  =  -- — 

51.  6ay-7ar2/  +  2?-2  =  0.  4  2  4 

52.  (3ax  +  4&)2  =  (2ax-&)l       54..  aV4-3aicV^-10Z»  =  0. 

55.  9A:2-cc2  =  2;ta;V3. 

56.  (aa;  +  fey^4-3(r^r  +  i)+2  =  0. 

57.  («V  -  3  ax)2  =  14(aV  -  3  aa^)  -  40. 


144  SECOND  COUESE  IN  ALGEBRA 

88.  Solution  by  formula.  In  Exercise  42,  above,  the 
general  quadratic  ax^  -{-hx-\-  c  =  Q  has  been  solved  and 
the  roots  found  to  be 

x  = (F) 

2a  ^    ^ 

The  expression  (i^)  is  a  general  result  and  may  be 
used  as  a  formula  to  solve  any  quadratic  equation  in  the 
standard  form  aa^  -\-  bx  -\-  c  =  0,  in  which  a,  b,  and  e  may 
represent  numbers,  single  letters,  binomials,  or  any  other 
form  of  algebraic  expression  not  involving  x. 

If  the  numbers  a,  b,  and  c  are  such  that  the  expression  6^  —  4  ac 
is  negative,  the  formula  contains  the  square  root  of  a  negative  num- 
ber, which  is  a  kind  of  number  not  yet  fully  considered  in  this  text. 
In  the  exercises  that  follow  it  will  be  assumed  that  only  such  numer- 
ical values  of  the  literal  coefficients  are  involved  as  will  not  make 
6^  —  4  ac  negative.  A  discussion  of  the  case  here  ruled  out  will  be 
found  in  Chapter  XII. 

EXERCISES 

Solve  for  x  by  formula  and  check  the  results  as  directed 
by  the  teacher : 

1.     4:X^-^SX  =   S. 

Solution.    Writing  in  standard  form, 

4ar2  +  8a:-3=0. 

Comparing  with  ax^  +  bx  +  c  =  0,  evidently  4  corresponds  to  a, 
8  to  6,  and  —  3  to  c.    Substituting  the  values  in  (F)  gives 


8±V64-4«4-(-3) 
2-4 


8±V64  -H48      -8±Vll2 


Check  as  usual. 


-8db4V7 
8 


8  8 


QUADKATIC  EQUATIONS  145 

2.  x"- ex -16  =  0.  9.  3£c  +  4  =  x\ 

3.  x^-x-l  =  0.  10.  Ua^  -Wx  =  26. 


2  -^        2" 


5.  4a;2  +  5a^-3.  12.  x  +  f  =  fa;2. 

e.2x-\-4.  =  x\  13.  f -i£c2z=2x.     • 

7.  X  =  1  -  xi  14.  6x  ==  1  +  2£c2 

8.  4^2  -  lliT  -  60  =  0.  15.  .^2  +  2cc  =  8  -  a-^ 
16.  3j9V^j9x  +  4. 

Solution.    Writing  in  standard  form, 

Here         a  =  3  j?^,     b=  —  p,     and     c=— 4. 
Substituting  these  values  in  the  formula  (F), 


^  -  (-p)±^(-py  -  4  •  Sp^(-  4) 
2.3jo2 


_  p±^p^  +  48 jp^  _  JP  =fc  7jp  _   4  -1 

Check  as  usual. 

17.  x''-2kx-Sk''  =  0.  „^    .^,,      13A;i      871- 

25.  6  h^  H 


18.  3a^  =  7ax  +  6x^. 

19.  4ic2  ^  A;a!  - 14/^2  =0.  ««     ,2  .    .      .   1 


26.  ic^  _^  ^>£c  +  -  =  0. 


20.  bx  =  12x^-b\ 

.2^2 


c 


21.  a'x^-\-4.abx-{-Sb^  =  0.        27.  -  +  a^  +  3^  =  0. 

22.  12pV-4j9ra;-?-  =  0. 

23.  Sa^x^-\-Sabx-h4.b^=0.     28.  —  -  -  _  i  =  0, 


24.  X  -  3  VJ  -  —  =  0. 


2 
ic  29.  13  a%^x^  =  9  ^/  -f  4  aV. 


146  SECOND  COURSE  IN  ALGEBRA 

30.  2x^  +  5x  =  cx^-{-Scx-\-S. 
Solution.      (2  -  c)  a;2  +  (5  -  3  c)  a;  -  3  =  0. 
Here      a  =  2  —  c,     6  =  5  —  3  c,     and     c  =  —  3. 


Hpi 

ace 

3  c  -  5  ±  V(5  -  3  c)2  +  12  (2  -  c) 

XlCJ 

2(2-c) 

3c-5±V49-42c  +  9c2 

4-2c 

_3c-5±(7-3c) 

4-2c 

2                       1               ^     6c-'l2 

4-2  c'     ^^     2-c'     "^     4-2c 

Check 

as  usual. 

31. 

P' 

-\-x''  =  22)x-j-2x-2p. 

32. 

(X 

-lf  =  a(x-x^). 

33. 

x'  +  2x  =  kx^-\-kx-l. 

34. 

Sx  +  ax^  =  2(x^  -{-ax-  1). 

35. 

cc^ 

—  ex  —  ^(ax  —  ac). 

36. 

4aV  +  4c2  =  16aV  +  cV. 

37. 

x« 

-a^x^=Sb'x^-Sa%\ 

38. 

a^ 

-i-x''  =  2ax-\-b\ 

39. 

12jpq  =  x^-}-4:qx-Spx. 

40. 

x^ 

-  aV  +  a%'  =  ^>*a;l 

or     -3. 


89.  Comparison  of  the  various  methods.    Four  methods 
have  been  given  for  the  solution  of  tlie  quacbatic  equation : 
(a)  Solution  by  factoring. 
(J)  Solution  by  graphing, 
(c)  Solution  by  completing  the  square, 
(c?)  Solution  by  formula. 
In  practice,  the  first  and  the  last  of  these  methods  are 
most  convenient. 


QUADRATIC  EQUATIONS  147 

When  an  equation  with  integral  coefficients  can  be 
solved  by  factoring,  as  explained  in  Chapter  III,  the  roots 
are  always  rational  numbers. 

For  example,  3x2  +  2a;-8  =  0  factors  into  (3  a;  -  4)  (a:  +  2)  =  0. 
Hence  the  roots  are  x  =  ^  and  —  2. 

An  inspection  of  the  formula 

-h±  Vb^  -4:ac 


2a 

shows  that,  if  a,  b,  and  c  are  integers,  one  always  obtains 
roots  involving  radicals  unless  the  expression  under  the 
radical  sign,  5^  —  4  ac,  is  a  perfect  square.  In  this  case, 
however,  the  values  of  the  roots  are  rational  numbers. 
For  example,  in  the  equation  3a;^  +  2a:  —  8  =  0  the  value  of  the 
expression  6^  —  4  ac  is  (2)^  —  4  •  3  (—  8)  =  100,  which  is  a  perfect 
square.  Hence  the  roots  of  3  a:^  +  2  a:  —  8  =  0  are  rational  num- 
bers, since  the  radical  term,  in  the  roots  can  be  expressed  as  a 
rational  number. 

Hence  to  determine  whether  a  quadratic  equation  of 
the  form  aa^  -j-bx-^  c  =  0  can  be  solved  by  factoring  into 
ratienal  factors,  we  have  the 

Rule.    Compute  the  value  of  b^  —  4:  ac  for  the  equation. 
If  the  result  is  the  square  of  an  integer,  the  left  member 
of  the  equation  can  be  factored  and  the  roots  are  rational. 

ORAL  EXERCISES 
Determine  which  of  the  following  can  be  solved  by  factoring  : 

1.  x"  +  lOcc  +  25  =  0.  6.  x^  -  4x  -  4  =  0. 

2.  x'-^x- 4.00  =  0.  7.  3 a;2  _  ic  +  4  =  0. 

3.  x2_7^_3  =  0.  '    8.  3a^2-5ic  +  8  =  0. 

4.  2x2_f_3^_^-^^Q  9^  2£c2-10£c-25  =  0. 

5.  2cc2  +  3x  +  2  =  0.  10.  3x^-h2ax-a^==0. 


148  SECOND  COUESE  IN  ALGEBRA 

REVIEW  EXERCISES 

Solve  the  following  by  the  method  best  adapted  to  each 

1.  ^x"  -7x  -  10  =  0.  6.  x'  -  7x2  +  12  =  q 

2.  5x2  +  14.r  +  8  =  0.  7.   6x«-20  =  2a;l 

3.  4x2  4-3x-2  =  0.  S.x^-x  =  0. 

4.  x^  -1.1  X-  .84  =  0.  9.  x^  =  x. 

5.  8x2-f-2V5x-15  =  0.      10.  x^  +  64x  =  0. 

11.  (a^-3x)'^-2(x2-3x)=8. 
12.  5^2- 9^  +  3-0.  ,_    x^-l      x^  + 1       _ 


x'  +  l 

14.  x2  +  xV2-V6  =  xV3.      19       14         X  -  1  ^  X  +  1 

a;  +  4x-2      x  +  2 


13.  .03x'  +  .01x  =  .l. 

x2  +  xV2-V 
X  —  1       X  -\-l 


^^'       X      -      6     '  ^^    2ic -1       1-x      x  +  1 

X  —  1        cc  —  2ic  +  2 

X  —  6  _    X  —  4: 

xHi  ~  2a; +  6*  ^1.  pqx^  -  rqx  -\- psx  =  rs. 

x±l      x  +  3^22  22.x'-2ax-\-a'~b'  =  0. 

24.  (x2  +  5x  +  2)2-6(«2-f5x  +  2)=16. 
3x-l      x  +  1  48 


25. 


5.^  +  1      x-1       (5x  +  l)(l-a:) 
a;»-10a^  +  l 


27.  15^2 -1.95x4- .054  =  0. 

28.  10-9x  =  7x''. 

29.  x^  +  961  a^  =  62  ax. 

30.  (x8  +  4)^  =  4  +  a:»: 

31.  (2x  +  3)(x-4)  =  (3x-8)(4x-l). 

32.  (3x  -  2)(x  -  5)  =  (4x  -  3)(x  +  1). 


QUADRATIC  EQUATIONS  149 

33. 

34. 

35. 

36. 

X  -j-  b  X  +  o 

38.  17a;  =  6ic2-10. 

39.  7x*-4t2x'-\-25  =  13x''-5x^-\-4.00. 

40.  2k^x^  -h  Skx  =  5(3kx  -1). 


1         2 

1 

x-^7      5 

3-a: 

^ 

a:  +  2 

ar^  +  Scc- 

4"^x  +  4 

£C  +  3 

x-1 

x-2 

x^-3x-\-2      x-2 

x-1 

5x-ll 

l-3a^ 

2x-l 

x-1      ' 

cc  +  2 

0^  +  1 

x  +  12      ^ 

,      x  +  7 

41. 

37j9^a:  =  210^2-pV. 

42. 

.x  +  1       ic-1                       2ic-f-l 

x^-1      x^  +  1       (x^^x  +  l)(x^- 

-0^  +  1) 

43. 

3x-2      2                5(^  +  1) 
0,-1     '3~^6-4^_i 

44. 

cc*-a;2^4(9  +  ^'). 

45. 

^   ,       4:k          k-2x 
e  —  Za3            c 

46. 
47 

4-30^        2x        2x-\-5 
X           1-x      3-2ic 

6cc2^x-5      12x=^H-2x-4 

4ic2-£c-2        8a^2-2£c  +  4 

48.  8.4x2  4- .005  a: -.15  =  0. 

49.  324  a;^  + 1936  =  1665x1 

50.  3x2  +  .7x=.2. 

51.  3 aV _  10 abx  -  25b^  =  2(2abx  -  aV). 


150  SECOND  COURSE  IN  ALGEBRA 

1  1  1 

52. ^ -7^  + 

53 


54. 


x'-Zx  +  2      x^  +  x-^      ?,-2x-x^ 

x  +  2 2  3(2-0;) 

(x  +  4.)(x-  1)       (x  +  4)(3  -x)~  {x-'d){l-x)' 
2(ar^-2)-l^       x'' +  2 
x^-1        ~(x'-2)-4.' 
55.   (3x  -  7)(2x  -^1)  -  (5x  -^  2)(2x  -  S)  =  0. 
.  56.   (x  -  2)2-  (1  -  2x)(Sx-\-  5)  =  5  -  (1  -  2x)(3a3  +  2). 

57.  (a32-2cc-2)2+2(cc2-2x-2)-3  =  0. 

58.  (£c2  -  4ic  +  1)2  _  4(cc2  -  4x)  -  16  =  0. 

59.  (2a;-^>)2  =  ci(2£c-Z>)  +  2a2. 

60.  «  +  -  =  w  H 

x  m 

PROBLEMS 

1.  Separate  42  into  two  parts  such  that  the  first  shall  be 
the  square  of  the  second. 

2.  Find  two  consecutive  odd  numbers  whose  product  is  143. 

3.  The  sum  of  the  reciprocals  of  two  consecutive  numbers 
is  \^.    Find  the  numbers. 

4.  The  time,  in  hours,  required  for  a  trip  of  216  miles  waa 
6  greater  than  the  rate,  in  miles  per  hour.  Find  the  time  and 
the  rate. 

5.  The  altitude  of  a  triangle  is  6  feet  less  than  the  base. 
The  area  is  bQ>  square  feet.    Find  the  base  and  altitude. 

6.  One  leg  of  a  right  triangle  is  9  feet  shorter  than  the 
other,  and  the  area  is  45  square  feet.    Find  the  three  sides. 

7.  The  area  of  a  trapezoid  is  180  square  feet.  One  base  is 
4  feet  greater  than  the  altitude,  the  other  is  three  times  the 
altitude.    Find  the  two  bases  and  the  altitude. 

8.  A  rectangle  whose  area  is  27  square  inches  has  a  perim- 
eter of  21  inches.    Find  the  length  of  the  shorter  side. 


QUADRATIC  EQUATIONS  151 

9.  A  polygon  of  n  sides  always  has  ^n(n  —  S)  diagonals. 
How  many  sides  has  a  polygon  with  119  diagonals  ? 

10.  If  AB,  in  the  accompany- 
ing figure,  is  a  tangent  to  the 
circle  and  BD  is  any  secant,  then 
(ABf  =  BC  .  BD.  li  AB  =  6  and 
CD  =  10,  find  BC. 

11.  A  requires  4  more  days  than 
B  to  do  a  piece  of  work.  Work- 
ing together  they  require  2|^  days. 
How  many  days  will  each  require 
alone  ? 

12.  In  selling  an  article  at  $20,  a  merchant's  per  cent  of 
profit  is  9  greater  than  the  original  cost  of  the  article  in 
dollars.    Find  the  original  cost  and  the  per  cent  of  profit. 

13.  A  3-inch  square  is  cut  from  each  corner  of  a  square 
piece  of  tin.  The  sides  are  then  turned  up  to  form  an  open 
box  of  volume  12  cubic  inches.  What  was  the  side  of  the 
original  square  ? 

14.  The  number  of  lines  on  a  certain  printed  page  is  20  less 
than  the  average  number  of  letters  per  line.  If  the  number  of 
letters  per  line  is  decreased  by  16,  the  number  of  lines  must 
be  increased  by  15  in  order  that  the  new  page  may  contain  as 
much  matter  as  the  old.  How  many  lines  and  how  many 
letters  were  there  on  the  original  page  ? 

15.  For  what  positive  value  or  values  of  x  will  the  product 
of  0?  —  5  and  a?  -f-  5  be  1  greater  than  their  difference  ? 

16.  What  values  of  x  will  make  the  product  of  4ic  —  5  and 
X  —  2  equal  in  value  to  sc  -f  4  ? 

17.  The  length  of  a  rectangular  room  exceeds  its  width  by 
2  feet.  A  rug  placed  in  the  middle  of  the  floor  leaves  a  mar- 
gin of  1  foot  all  around.  If  the  area  of  the  rug  is  eight  times 
that  of  the  margin,  find  the  dimensions  of  the  room. 


152  SECOND  COURSE  IN  ALGEBRA 

18.  A  sum  of  |4000  is  invested,  and  at  the  end  of  .each  year 
the  year's  interest,  plus  |400,  is  added  to  the  investment.  At 
the  beginning  of  the  third  year  the  investment  afnounts  to 
$5230.    What  is  the  rate  of  interest? 

19.  The  cost  of  an  outing  was  $60.  If  there  had  been  two 
more  in  the  party,  the  share  of  each  would  have  been  a  dollar 
less.    How  many  were  there  ? 

20.  The  dimensions  of  a  rectangular  box  are  expressed  by 
three  consecutive  numbers.  Ics  surface  is  214  square  inches. 
Find  its  dimensions. 

21.  The  radius  of  a  circle  is  28  inches.  How  much  must  this 
be  shortened  in  order  to  decrease  the  area  of  the  circle  by  1078 
square  inches  ? 

22.  Two  bodies,  A  and  B,  move  on  the  sides  of  a  right 
triangle.  A  is  now  5  feet  from  the  vertex,  and  moving  from 
it  at  the  rate  of  10  feet  per  second.  B  is  35  feet  from  the  ver- 
tex, and  moving  toward  it  at  5  feet  per  second.  At  what  time 
(past  or  future)  are  they  75  feet  apart  ? 

23.  How  high  is  a  mountain  which  can  just  be  seen  from  a 
point  on  the  surface  of  the  sea,  40  miles  distant  ? 

Hint.    See  Exercise  10. 

24.  The  distance  in  feet,  s,  through  which  a  body  fallg  from 
rest  in  t  seconds,  is  given  by  the  formula  s  =  I  gf/^,  where 
<7  =  32,  approximately.  A  bomb  dropped  from  an  aeroplane 
struck  the  ground  below  8  seconds  later.  How  high  was  the 
aeroplane  at  the  time  ? 

25.  If  a  stone  be  thrown  vertically  upward,  with  a  velocity 
of  100  feet  per  second,  it  is  known  that,  neglecting  the 
resistance  Of  the  air,  its  height  after  t  seconds  will  be 
100^  —  16^^  feet.  After  how  many  seconds  will  it  be  136  feet 
high?    Explain  the  double  answer. 

26.  After  how  many  seconds  will  the  stone  of  Exercise  25 
return  to  its  starting  point  ?   Explain  the  zero  root. 


CHAPTER  XI 

IRRATIONAL  EQUATIONS 

90.  Definitions  and  discussion.  An  irrational  equation  in 
one  unknown  is  an  equation  in  which  tlie  unknown  occurs 
under  a  radical,  or  is  affected  by  a  fractional  exponent. 

Thus  3x  -\-5Vx  =  l,  x-x^  +  1  =  0,  and  Vx^  -3x4-2  =  6  are 
irrational  equations. 

The  chief  difficulty  involved  in  the  solution  of  sucK 
equations  arises  from  the  fact  that  sometimes  results  are 
obtained  which  do  not  satisfy  the  given  equation  and 
hence  are  not  roots  of  that  equation.  A  result  of  this 
kind  is  called  extraneous. 

EXAMPLE 


(a)  Solve  Vic  —  6  —  4  = 

:0. 

(^) 

Solve  -Vcc-6-4: 

=  0. 

Solution.    Transposing, 

Vx  -  6  =  4. 

(1) 

-Va:-6  =4. 

(1) 

Squaring,          a;  —  6  =  16. 

(2) 

a:  -  6  =  16. 

(2) 

Solving,                    X  =  22. 

X  =  22. 

Check.  V22  -6-4  =  0. 

-  V22  -6-4  =  0. 

Vl6  -4  =  0. 

-Vi6-4  =  0. 

4-4  =  0, 

-4-4  =  0, 

lich  is  true. 

which  is  not  true. 

It  appears  from  a  study  of  these  solutions  that  state- 
ments (1)  differ  only  in  the  signs  preceding  their  left 
members.     Consequently  this  distinction   disappears  after 

»K  163 


154  SECOND  COUESE  IN  ALGEBRA 

squaring,  and  equations  (2)  are  identical.  Since  the  re- 
mainder of  the  work  in  both  («)  and  (5)  consists  in  the 
solution  of  (2),  the  result  obtained  is  really  the  root  of 
this  equation.  Whether  the  root  obtained  satisfies  both 
(a)  and  (5),  or  only  one  of  them,  can  be  determined  only 
by  substitution.  Hence  it  appears  that  (a)  is  an  equation 
and  that  (5)  is  not,  but  is  merely  a  false  statement  in  the 
form  of  an  equation. 

In  any  case,  all  of  the  roots  of  the  original  equation  are 
sure  to  be  among  the  results  found,  provided  no  factor 
containing  the  unknown  has  been  divided  out.  But  no 
result  should  be  called  a  root  unless  it  satisfies  the  origiual 
equation.    This  means  that  all  results  must  be  checked. 

In  irrational  equations,  as  m  all  the  work  up  to  the 
present,  it  is  understood  that  unless  a  radical  or  an  ex- 
pression affected  by  a  fractional  exponent  is  preceded  by 
the  double  sign  ±  it  has  only  the  one  value,  just  like  any 
other  number  symbol. 

Thus  Vl6  means  +  4,  and  not  —  4. 

Also  4^  means  +  2,  while  —  ^^  means  —  Vi,  or  —  2. 

.  If  this  fact  is  kept  in  mind,  it  is  clear  from  an  inspec- 
tion of  (5),  above,  that  it  could  have  no  root,  since  the 
sum  of  two  negative  numbers  could  not  possibly  be  zero. 
The  method  of  solving  equations  containing   radicals 
is  stated  in  the 

Rule.  Transpose  the  terms  so  that  one  radical  expression 
(the  least  simple  one  if  there  are  two  or  more}  is  the  only 
term  in  one  member  of  the  equation. 

Next  raise  both  members  of  the  resulting  equation  to  the 
same  power  as  the  index  of  this  radical. 

Combine  like  terms  in  each  member  and^  if  radical  ex- 
pressions still  remain^  repeat  the  two  preceding  operations 


lERATIONAL  EQUATIONS  155 

until  an  equation  is  obtained  which  is  free  from  radicals ; 
then  solve,  this  equation. 

Check.  Substitute  in  the  original  equation  the  values  found 
and  reduce  the  resulting  numerical  equation  to  its  simplest 
form  by  extracting  roots,  but  not  by  raising  both  members  of 
the  equation  to  any  power. 

Finally,  reject  all  extraneous  roots. 

EXERCISES 

Solve  the  following  for  real  roots  and  check  results : 
1.   Vcc  +  1  =  5.  7.   Vic  +  l=V3x~5. 

2..   -s/?>x-5  =  7.  •  8.  2V8x  =  xVi. 


3.  ■\/2a^-h3  =  3.  9.  3  V2a:  +  6  =  VGx^  -  6. 

4.  2  Vx  +  4  =  4.  10.   -\/4x  +  3--v/4-3a^  =  a 


5.  3V2x-8-7  =  17.  11.  V2x  +  2=V2x-2. 

6.  7  +  2a/2cc-1=:13.  12.   ^iTl=V^^^. 

13.  1+Vcc-2=V^. 
Solution,    Transposing, 
Squaring, 

Transposing  and  collecting, 
Squaring, 
Check.    Substituting  |  for  x  in  the  original  equation, 

l  +  V|^=V|, 

14.  V9ic-  4  =  Vcc. 

15.  V3ic-2  +  5  =  Vcc4- 35. 

16.  V3  .'^5  +  4  —  Vcc  +  5  =  Vd  —  X. 


17.  3Va;  +  4  + Vl-.T==3Vl-ic  +  2Vic  +  4. 

18.  Va;  +  2  =  Vi  +  V2.  19.  a;  -f  1  =  V^ic^  +  S; 


156  SECOND  COUKSE  IN  ALGEBKA 

20.  ir -h  4  +  V^  =  V^M^e. 


21.   V2x-4  4-Va;  +  6  =  VT^ 


22.   -Wx  -\-  2  -\-  -Vx  -1-  ^3  X  -\-  3  =  0. 


23.  V3cc+1- Vic+1=  Vx-4. 

24.  x^- 9x^  +  20  =  0. 

Note.  The  equation  here  given  is  not  a  quadratic  equation,  but 
it  is  of  the  general  type  ax^^  +  6a;'»  +  c  =  0.  Here  x  occurs  in  but 
two  terms,  and  its  exponent  in  one  term  is  twice  that  in  the  other 
term.  Many  equations  in  this  form  can  be  solved  by  completing  the 
square  (compare  Exercise  42,  p.  143). 

Solution.  x^  -  9  xt  +  -\L  =  _  20  +  -S|^. 

art  =  5  or  4. 
Whence  a;  =  ±  5*  or  ±8. 

Check.    Substituting  ±  5^  for  x  in  the  original  equation, 

(±5f)|_9(±  51)1  +  20  =  0, 
or  52  -  9  .  5  +  20  =  0,  or  0  =  0. 

Substituting  ±  8  for  a:  in  the  original  equation, 

(±  8)i-9(±  8)^  +  20  =  0. 

16  -  36  +  20  =  0,  or  0  =  0. 

25.  x^  +  5a;'  -14  =  0.  30.  2a;^  -  7  -^sc  +  6  =  0. 

26.  .'-10x1=11.  31...-!  =  4x-i  +  32. 

,  32.  aji  -  a;i  -  6  =  0. 

27.  6  =  x^-{-x.  OK         1 

33     r-i  —  -4-  i  —  0 

28.  Sxi  -{-5x^-\-2  =  0.  SQxi      9~^' 

29.  x-''-17x-^-\-52  =  0.         34.  7ic^-6=2a;i 

35.  x^-^5x  +  aVa^^  +  Sx  -  54  =  0. 
Hint.   Let  y  =  x^  ■}■  6x. 


lEEATIOKAL  EQUATIONS  157 


36.  3x2-4x-llV3ic^- 

4a; +28=0. 

37.  2x^-'^x-^--\/2x'- 

-3a^-l+l  =  0. 

38.  x^-lx-h-^x"-  -2x- 

-4  +  2  =  0. 

39.   V2x-4  +  5  =  l. 

40.  <Ix^?>=^^\x-X 

41.  12£c^-27ic'^  =  20£c3  _ 

-45. 

42.  15-2Vir2_^9  =  x^  +  9. 

43.  27(8  +  27a;*-ic^)  =  8xi 

V2a^-V8 

44.  4Vic  +  2  =  3Vic  +  4. 
45.-  ^x  +  4  =  Va^  -  8. 

2V3          \/2a:  +  12 
V2x-3           3V3 

48.  Vx  +  1  =  Va;  -  4  +  1. 

49.  4(cc  -  5)  -  2(aj  _  5)*  -  2  =  0. 

50.  ^x^'x  =  {x  -\-  3) V9^. 

^,  Va;  +  2-     ^8-^       8 

51.  ,  —     ,  =  o  * 

V8  -  cc      Vx  +  2      3 

52.  5x^  +  «7x^  -  200  =  16  -  3icl 

53.  4  V^- 21  ^^  +  27  =  0. 

54.  ix"  -  5a;  +  2)*  -  h{x?  -  5cc  +  2)^  +  6  =  0. 

55.  V  17  +  2  V3  +  s  +  V7+?-  5  =  0. 


56.  3cc2-a;-22  =  6V3ic'-cc- 6. 

57.  If  #  =  TT-v  -J  solve  for  I  and  cr. 

>^ 

58 .  VtT  =  \x  V3  and  ^  =  2 7rr^,  express  A  in  terms  of  x. 

59.  If  ^^•=  2  7'^  and  C  =  2  vrr,  express  C*  in  terms  of  a. 

60.  If  ^  =  — -  V  3  and  a;  =  -|-  r  V3,  express  A  in  terms  of  x. 

61 .  If  i!L  =  3  ?-^  and  a  =  J  ?■  v  3  +  V2,  express  i^  in  terms  of  a. 


CHAPTER  XII 

IMAGINARIES 

91.  Definitions.  When  the  square  root  of  a  negative 
number  arose  in  our  previous  work,  it  was  called  an 
imaginary,  but  no  attempt  was  then  made  to  use  it  or  to 
explain  its  meaning.  The  treatment  of  imaginaries  was 
deferred  because  there  were  so  jnany  topics  of  more  im- 
portance to  the  beginner.  It  must  not  be  supposed, 
however,  that  imaginaries  are  not  of  great  value  in  mathe- 
matics. They  are  frequently  used  in  certain  branches  of 
applied  science ;  and  it  is  unfortunate  that  symbols  which 
can  be  employed  in  numerical  computations  to  obtain  prac- 
tical results  should  ever  have  been  called  imagmary.  By 
such  a  name  something  unreal  and  fanciful  is  suggested, 
to  obviate  which  it  has  been  proposed  to  call  imaginary  num- 
bers orthotomic  numbers^  but  this  name  has  been  httle  used. 

The  equation  a:^  + 1  =  0,  or  x^  =  —  l^  states  that  a:;  is  a 
number  whose  square  is  —1.  By  defining  a  new  number, 
V— 1,  as  one  whose  square  is  —1,  we  obtain  one  root  for 
the  equation  x^  +  l  =  0. 

Similarly,  V—  5  is  a  number  whose  square  is  —  5.  And, 
in  general,  V—  rv  is  a  number  whose  square  is  —  n.  Obvi- 
ously, V—  6  means  something  very  different  from  V5, 
and  V—  n  from  Vn. 

The  positive  numbers  are  all  multiples  of  the  unit  -f  1, 
and  the  negative  numbers  are  all  multiples  of  the  unit  —  1. 
Similarly,  pure  imaginary  numbers  are  real  multiples  of 
the  imaginary  unit  V— 1,  as  2V— 1,  —  5V— 1,"  and  5V— 1. 

168 


IMAGINARIES  159 


Furthermore  v  —  4  =  Sv—  1 ;  -y  —  a^  =  a^/—  1 ;  and 
V^=V5.V^.         __ 

The  imaginary  unit  V—  1  is  often  denoted  by  the  letter 
I ;  that  is,  3  V—  1  =  3  z. 

If  a  real  number  be  united  to  a  pure  imaginary  by  a 
plus  sign  or  a  minus  sign,  the  expression  thus  obtained  is 
called  a  complex  number. 

Thus  —  2  +  V— 1  and  3  —  2V—  4  are  complex  numbers.  The 
general  form  of  a  complex  number  is  a  +  hi,  in  which  a  and  h  may 
be  any  real  numbers. 

Note.  Up  to  the  time  of  Gauss  (1777-1855)  complex  numbers 
were  not  clearly  understood  and  were  usually  thought  of  as  absurd. 
The  situation  reminds  one  of  the  time  when  negative  numbers  were 
similarly  regarded,  and  the  veil  was  removed  from  both  in  about  the 
same  way.  It  was  found  that  negative  numbers  rfeally  had  a  sig- 
nificance —  that  they  could  be  used  in  problems  that  involve  debt, 
opposite  directions,  and  many  other  everyday  relations.  The  inter- 
pretation of  imaginary  numbers  is  not  quite  so  obvious,  and  is  not 
considered  in  this  text.  But  as  soon  as  it  was  seen  that  an  interpre- 
tation was  possible  the  ice  was  broken,  and  it  needed  only  the  insight 
and  authority  of  a  man  like  Gauss  to  give  complex  numbers  their 
proper  place  in  mathematics. 

ORAL  EXERCISES 

Express  as  multiples  of  V— 1,  or  i : 


1.  V-16. 

5.  3  V-  6. 

2.  V-25. 

6.  2^-c. 

3.  V-6^2_ 

7.  V3.V-3. 

4.  2V-3. 

8.   V2.V-5. 

11.   ^-x'-2x-l. 


12.   V-?/  +  62/-9. 

92.  Addition  and  subtraction  of  imaginaries.  The  funda- 
mental operations  of  addition  and  subtraction  are  performed 
on  imaginary  and  complex  numbers  as  they  are  performed 
on  rational  numbers  and  ordinary  radicals  of  the  same  form. 


160  SECOND  COURSE  IN  ALGEBKA 


Thus 
and  5V^-3V^=2V^. 

Also    (8  +  5V^)+(4-2V^)  =  7+3V^. 
Similarly,  (a  +  bi)  +  (c  +  di)  =  a -{■  c  ■\- (b  +  d)  i 


Simplify : 

1. 3V^+2Viri. 

2.  4V^4-V^. 


EXERCISES 


7.   V-18  +  V^. 


3.  V-  25  -  V-  16. 

4.  V^  +  V^. 


5.  V-4  4-V~16. 

6.  (_  8)^  +  (- 32)^. 


8.  4V-25cc2_2V-36ar». 

9.  24-3V^  +  6-5V^. 

10.  7VI^2-5a  +  4V^ir;^. 

11.  (Sa-  6ib)-]-(a-\-{h). 

12.  4-8i  +  16-  3V^. 


13.  5  4-3V-49cc2_6V^6x2H-4. 
14.^  18  -  3(-  1)*  +  6(-  25)*  +  4. 

15.  5  V^  +  3  V^  _  V-  27  +  2  V^s. 

16.  (8-5  V^Il6)-(7  +  3  V-  25). 


17.  4  V-  9  a*  -  6  a^  V- 16  +  3 

18.  (5x-6iy)-(Sx-\-2iy). 


6  +  5  V-  54. 


93.  Multiplication  of  imaginaries.    By  the  definition  of 
square  root,  the  square  of  (—  ny  is  —  n. 

Therefore     {y/'^f  =  -l, 

(V3i)3  =  (V3i  )VZi  =  _  V3T. 

(v::i  )^ = (vzi  )2  ( vzT)2 = (_  1 )  (_  1) = 1. 


IMAGIKARIES 


161 


'^) 


To  multiply  V—  2  by  V—  3,  we  first  write 

v32'^V2.viri, 

and  V^  =  V3.V3l. 

Then       V32  .  V^g  =  ( V2  .  V:^) ( Vs  •  VI3) 

==  V6  .  V3i  .  V^=-V6. 
Similarly, 

(2 Vi:5)(-  3 Vr2)=  (2V5  .  vC3)  .  (-  3V2  •  a 

=  _6VlO(-l)  =  6VlO. 

In  general,  if  V^^  and  V— 5  are  two  imaginaries  whose 
product  is  desired,  they  should  first  be  written  in  the  form 
Va  .  V—  1  and  y/b  •  V—  1  and  the  multiplication  should  only 
then  be  performed.    This  method  will  prevent  many  errors. 

In  this  connection  it  must  be  clearly  understood  that  one  rule 
followed  in  the  multiplication  of  real  radicals  (see  page  117)  does 
not  apply  to  imaginary  numbers. 

In  the  case  of  ordinary  radicals  we  have 


But  the  product  of  two  imaginaries  like  V— 2  •  V—  3  does  not 
equal  V(—  2)(—  3),  for  this  equals  Vc.    We  have  seen  above  that 

In  multiplying  two   complex  numbers,  write  each  ex- 
pression in  the  form  a  ±  hi  and  proceed  as  in  the  following 


EXAMPLE 

Multiply  4  +  V^  by  2  -  V^. 
Solution.  4  +  V^  =  4  +  V5  .  V^, 

2  _  VITe  =  2  -  Ve .  V^. 


(1) 
12) 


Multiplying  (1)  by  (2),  8  +  2  Vs  •  V^  -  4  V6  •  V^-  V30  (-  1). 
Rewriting,  8  +  2  V^  -  4  V^  +  V30. 


162 


SECOND  COUKSE  IN  ALGEBRA 


EXERCISES 

Perform  the  following  Indicated  multiplications  and  simplify 
results : 


17.   Vm  +  n  '  V—  7n  —  n. 

18.  (3+v:ri)(3-V:^). 

19.  (5+V^)(5-V^). 

20,  (3-4V2^)(3  +  2V2^). 

21.  (4+v:^)(5-V:r3).. 

22.  (5-3i)(6-5V2i). 

23.  {a  +  ib){c  +  id). 

32.  (2  +  2  aAITs)'  -  (2  -  2  V^ 


25.  {a-\-ib)(a-ib). 


26.  (-i  +  iVirs)^ 

27.(-i-iV33;. 

28.  (-2  4-2V^)'. 

29.  (-2-2V^)'. 

30.  (x  —  iy)\ 

31.  {a  +  ibf-{a-ibf. 


Sf. 


33.  (a  +  ^  Vl  -  b^)  {a  -  I  Vl  -  h'^). 

34.  Determine  whether  the  sum  and  the  product  of  the 
numbers  5  +  6  V—  2  and  5  —  6  V—  2  are  real. 

35.  Determine  whether  the  sum  and  the  product  of  the 
numbers  2  +  V—  3  and  2  —  V—  3  are  real. 

94.  Division  of  imaginaries.  One  complex  number  is 
the  conjugate  of  another  if  their  product  and  their  sum 
are  real.    Thus  a  +  hi  and  a  —  hi  are  conjugates.   Conjugate 


IMAGINAEIES 


163 


complex  numbers  are  used  in  division  of  imaginary  ex- 
pressions as  conjugate  radicals  are  used  in  division  of  real 
radicals. 

In  case  either  the  numerator  or  the  denominator  of  a 
fraction  is  imaginary  or  complex,  the  division  may  be 
performed  as  in  the  following 


1.   V^-v-V2. 


EXAMPLES 


V-6.V2       2V-3 


OVIUUIUU. 

V2            (V2)^               2 

2.   V8--V^ 

^. 

Solution. 

V8        V8.V-2       4ti 

Vi:2     (V^)'     -  2 

-21 

3.   V-6--^ 

7:^2. 

Solution. 

V-  ()      V-  6  .  V-  2      V6  • 

'•^^•'_,/3 

v:^      C^/^)' 

-2              ^^• 

4.  3-(2+- 

s/33). 

Ortll-l<-Trt« 

3          _            3  (2  -  V-  3) 

6-3V-3 

2  + 

V-3       (2  +  V-3)(2-V- 
6  -  3  V- 3 

-3)            4  +  3 

The  method  of  the  above  examples  is  stated  in  the 

Rule.  Write  the  dividend  over  the  divisor  in  the  form  of 
a  fraction. 

Then  multiply  both  numerator  and  denominator  of  this 
fraction  hy  the  simplest  expression  which  will  make  the  new 
denominator  real  and  rational. 

Reduce  the  result  to  its  simplest  form. 


164 


SECOND  COUESE  IN  ALGEBRA 


1. 

V-  12  . 

-V2. 

2. 

Ve^v 

^-2. 

3. 

2VE^ 

4V-1. 

4. 

V-9^ 

-V-1. 

5. 

i^Vi: 

~3. 

6. 

4-^V- 

"5. 

7. 

Vs^v 

'-2. 

8. 

(-  49)* 

^(-64)*. 

9. 

^by^^ 

^-y. 

10. 

V—  m  - 

f-V—  7i. 

21. 

.  (3  +  2i)(l 

EXERCISES 

Perform  the  indicated  operations  : 

11.  (-  6/>a;)*-f-(-5a;)t 

12.  [(-a^6■)i_(-_^2^^i]^(^_^)i 

13.  3-(l+V^). 

14.  2-(l-V^). 

15.  2V^^(V^H-6). 

16.  2V^-f-(3V^  +  3). 

17.  (_i_v^)^(-i+v:r3). 

18.  (l+2^)--(3-4^). 

19.  X  -r-(£C  +  ^^/). 

20.  {(i-^ih)^{c-^id). 


22.  Is  4- 1  -  V-  3  a  cube  root  of  -  8  ? 

23.  Does  a;2_6ic4-12  =  0ific  =  3±  V^^s  ? 

24.  Does  X  =  f(V-lO),  y=-  |-(V-10),  satisfy  the  sys- 
tem x'-xy-  12  2/-2  =  8,  x"  +  xy-10y^  =  20  ? 

95.  Equations  with  imaginary  roots.  The  student  should 
now  be  able  to  check  the  solution  of  an  equation  which 
has  imaginary  roots. 

EXERCISES 

Solve  the  equations  which  follow,  and  check  the  results : 

1.  x'  +  4a;  +  8  =  0.  6.  cc^  +  a;  +  1  =  0. 

2.  3,2  _  8a;  +  24  =  0.  7.  oi^  -  a;  +  1  =  0. 

3.  ic2  +  3a;  +  9  =  0.  8.  5a;=^  -  6a;  +  14  =  0. 

4.  a;2-6a; +16  =  0.  9.  6 a;^  +  10 a;  +  21  =  0. 

5.  3a;2  4-2a;  +  4  =  0.  10.  3ar^  +  16a;  +  21  =  0. 


IMAGINARIES  165 

11.  x^  =  l. 

Hints.  If  x^  =  1,                          x^  -  1  =  0. 

Hence  (x  -  1)  (x^  +  x  +  1)  =  0. 

Then  .                         x  -  1  =  0, 

and  x2  +  X  +  1  =  0. 

12.  ic^  +  1  =  0.  15.  x^  =  l.  18.  x^  =  64. 

13.  x^  =  8.  16.  x^  =  9.  19.  x^  =  64. 
U.,x^=-27.                17.  £c'  =  l.              20.  x^  =  -125. 

21.  How  many  square  roots  has  any  real  number?  cube 
roots  ?    fourth,  roots  ?    sixth  roots  ? 

22.  What  do  the  preceding  exercises  suggest  regarding  the 
number  of  Tith  roots  which  any  real  number  has  ? 

23.  27 ic^- 8  =  0.  27.  a^«  +  7a;^-8  =  0. 

24.  64a;«f  125  =  0.  *  28.  3ic*  +  16cc2  +  21  =  0. 

25.  x''  -  2ic2  -.8  =  0.  29.  27  o;^  -I2x''- 64.  =  0. 

26.  x^-\-x^-2x-2  =  0.  30.  6x^  +  21x^  +  9  =  0. 

31.  25£c^  +  40a^2_^64  =  0. 

32.  (x^  4-  4)(.x2  4-  3  a;  +  7)  =  0. 

33.  (^2  +  20^)2  +  15  (x^  +  2a:)  +  54  =  0. 

34.  (x^  +  5xy  +  9(x^  +  5x)-  112  =  0. 

35.  Solve  a:  +  ?/  =  4,  x^  —  3  xi/  —  y^  =  —  39,  and  check  the 
results. 

36.  Solve  X  -\-  2  y  =  4z,  'if  —  X  =  0,  and  check  the  results. 

37.  Solve  x^  -\-  if  =  4:,  X  —  1/  =  6,  and  check  the  results. 

Note.  Long  before  the  time  of  Gauss,  mathematicians  had 
performed  the  operations  of  multiplication  and  division  on  com- 
plex numbers  by  the  same  rules  that  they  used  for  real  numbers. 
As  early  as  1545  Cardan  showed  that  the  product  of  5  -f  V— 15 
and  5  —  V— 15  was  the  real  number  40.  However,  he  was  not 
always  equally  fortunate  in  obtaining  correct  results,  for  in  another 

place  he  sets  -(  —  \  —  -)  =  — — =  =  -• 
4\      >J     4/      V64      8 


166  SECOND  COURSE  IN  ALGEBRA 

Even  the  rather  complicated  formula  for  extracting  any  root  of 
a  complex  number  was  discovered  in  the  early  part  of  the  eighteenth 
century.  But  all  these  operations  were  purely  formal,  and  seemed 
to  most  mathematicians  a  mere  juggling  with  symbols  until  Gauss 
showed  clearly  the  place  and  usefulness  of  such  numbers. 

96.  Factors  involving  imaginaries.  After  studying  radi- 
cals we  enlarged  our  previous  notion  of  a  factor  and,  with 
certain  limitations,  employed  radicals  among  the  terms  of 
a  factor.  Now  in  a  similar  manner,  with  like  restrictions, 
we  extend  our  notion  of  a  factor  still  farther  and  use 
imaginary  numbers  as  coefficients  or  as  terms  in  a  factor. 
For  example,  x^  +  1  may  hereafter  be  regarded  as  factorable, 

for  a^-\-l=2^-(-l)  =  {x+V^){x-^'^). 

Similarly, 

4  :^^  +  9  =  4  2^2  _  (_  9)  _  (2  ^^  +  3  V^)  {2x  -  3  V^) 

and  a:2  4-  6  =  a:2  -  (-  6)  =  {x  +  V^^)(2:  -  V^). 

Further,  a^  -1=  (^x  -1}  (^x^  -{-  x  -hi) .  Hitherto  the  trino 
mial  a:^-{-x-\-l  has  been  regarded  as  prime ;  but  the  stu- 
dent can  easily  prove  that  a^  +  x+lis  equal  to  the  product 
(2;  +  l4.jV33)(a:  +  | -^V^).  Therefore  a^-1  has 
three  factors,  these  two  and  x  —  1. 

Note  on  the  use  of  imaginaries.  We  have  explained  the  laws 
of  addition,  subtraction,  multiplication,  and  division  for  imaginary 
(and  complex)  numbers  and  have  made  some  use  of  them.  It  is 
largely  because  imaginaries  obey  these  laws  that  we  call  them 
numbers,  for  it  must  be  admitted  that  we  cannot  count  objects 
with  imaginary  numbers.  Nor  can  we  state  by  means  of  them  our 
age,  our  weight,  or  the  area  of  the  earth's  surface.  It  should  be 
remembered,  however,  that  we  can  do  none  of  these  things  with 
negative  numbers.  We  may  have  a  group  of  objects  —  books,  for 
example  —  whose  number  is  5  ;  but  no  group  of  objects  exists  whose 
number  is  —  5,  or  —  8,  or  any  negative  number  -wliatever.  If  it  is 
asked,  How,  then,  can  negative  numbers  and  imaginary  numbers 


IMAGINARIES  167 

have  any  practical  use  ?  the  answer  is  this :  They  have  a  practical 
use  because  when  they  enter  into  our  calculations  and  we  have 
performed  the  necessary  operations  upon  them  and  obtained  our 
final  result,  that  result  can  frequently  be  interpreted  as  a  concrete 
number  like  those  dealt  with  in  ordinary  arithmetic.  Moreover,  if 
the  result  cannot  be  so  interpreted,  it  is,  in  applied  mathematics 
at  least,  finally  rejected. 

In  that  part  of  electrical  engineering  where  the  theory  and  meas- 
■  urement  of  alternating  currents  of  electricity  are  treated,  complex 
numbers  have  had  extensive  use.  Their  employment  in  the  difficult 
problems  which  there  arise  has  given  a  briefer,  a  more  direct,  and  a 
more  general  treatment  than  the  earlier  ones  where  such  numbers  are 
not  used. 

In  theoretical  mathematics  complex  numbers  have  been  of  great 
value  in  many  ways.  For  example,  numerous  important  theorems 
aboat  functions  are  more  easily  proved  under  the  assumption  that 
the  variable  is  complex.  Then,  by  letting  the  imaginary  part  of  the 
complex  number  become  zero,  we  obtain  the  proof  of  the  theorem 
for  real  values  of  the  variable.  Indeed,  the  student  need  not  go  very 
far  beyond  this  point  in  his  mathematical  work  to  learn  that,  if  e  is 
2.7182  +  (see  page  253),  e  ""^l^  +  e-"^/-i  is  equal  to  the  real  number 
1.082  +  •  At  the  same  time  he  will  learn  also  how  such  a  form  arises, 
and  something  of  its  importance.  In  a  way  which  we  cannot  now- 
explain,  even  so  involved  an  expression  as  (a  +  16)''  +  *'' has  in  higher 
work  a  meaning  and  a  use.  If  the  student  pursues  his  mathematical 
studies  far  enough,  that  meaning  and  use  and  a  multitude  of  other 
uses  for  complex  numbers  will  become  familiar  to  him.  But  the 
numbers  which  we  have  learned  in  this  book  to  use,  namely  fractions, 
negative  numbers,  irrational  numbers,  and  compl-ex  numbers,  com- 
plete the  number  system  of  ordinary  algebra,  for  it  can  be  proved 
that  from  the  fundamental  operations  no  other  forms  of  number 
can  arise. 


CHAPTER  XIII 

THEORY  OF  QUADRATIC  EQUATIONS 

97.  Formation  of  equations  with  given  roots.  According 
to  section  34,  the  equation  (x—2)(x—S)=0  has  the  roots 
2  and  3.  In  general,  the  equation  (a;  —  r^)  (a;  —  rg)  =  0  has 
the  roots  r^  and  r^,  because  either  of  these  numbers, 
when  substituted  for  x,  satisfies  the  equation.  Hence  we 
can  always  find  an  equation  whose  roots  are  two  given 
numbers  r^  and  r^  by  setting  the  product  of  the  binomials 
x  —  Ti  and  x  —  r2  equal  to  zero. 

For  example,  an  equation  whose  roots  are  3  and  4  is  seen  in 
(a;  -  3)  (x  -  4)  =  0,  or  a;^  -  7a;  +  12  =  0. 

EXERCISES 
Form  an  equation  whose  roots  are  the  following : 

1.  2,3.  5.  2,|.  9.  1  +  V3,  1-V3 

2.  3,  7.  6.  5,  f  10.  3  ±  V7. 

3.  1,  -  3.  7.  -  I,  1 .  11.  2  +  V^,  2  -  V^. 
4.-2,-5.     is.  -|,-|.       12.   -7+V:r5, -7-V^. 

13.  i+Vf,i-\/|.  17.  1,-1,2. 

14.  i+vr|,i_vri.      i8..i,i,.i.      _ 

15.  2,  3,  4.  19.   Vir2,  -  V-  2,  2. 
Hint,    (x  -  2)  (x  -  3)  (x-  4)  =  0.         20.    1  +  V2,  1  -  V2,  3. 

16.  1,  3,  5.  21.1,- 1,  V^,  ~  VI^i. 

22.  V2  -  V3,  V2  +  V3,  -  V2  -  V3,  -  V2  +  Vs. 

168 


THEORY  OF  QUADRATIC  EQUATIONS        169 

98.  Relations  between  roots  and  coefficients.  By  direct 
multiplication  we  obtain  from 

(x-r^)(x-r,)=0  (1) 

the  equation        x^  —  (r-^  +  r^)  x  +  r^r^  =  0.  (2) 

Since  r-^  and  r^  are  the  roots  of  (1),  it  appears  from  an 
inspection  of  (2)  that  the  quadratic  equation 

x^-\-hx-\-c=0 

has  the  roots  r-^  and  r^^  provided  h  =  —  (r-^-\- r^)  and  c  =  r^r^. 

For  example,  we  may  form  at  once  the  equation  whose  roots  are 
4  and  9,  as  follows  : 

x-2-(4  +  9)x  +  4  •  9  =  0,  or  a:2  _  13  a;  +  36  =  0. 

Similarly  for  the  cubic  equation  x^  +  bx^  +  ex  +  d  =  0,  whose 
roots  are  r^,  r^,  and  r^,  we  have  (x  —  r^)(x  —  r^)  (x  —  r^)  =  0. 

Then  b  =-(r^ -{- r.-, -\-  r^), 

c  =  r^r^  +  r^r^  +  r^r^, 
and  d=—  r^r^r^. 

ORAL  EXERCISES 

Form  equations  whose  roots  are  the  following : 

1.2,9.  4.-3,-5.  7.  2V2,  -2V2. 

•    2.  4,5.  5.  -7,2.  8.  3  +  Vt,  3  -  Vz. 

3.  -1,6.  6.   V3,  -V3.  9.  I+V2,  I-V2. 

We  will  now  show  the  relations  which  exist  between  the 
roots  and  the  coefficients  of  the  general  quadratic  equation 
a2^+  hx-\-  c=Q.  By  section  88  the  roots  of  ax^  -\-hx+  c  =  0 
are 


and     r«  = . 

2a  2  2a 


170  SECOND  COUKSE  IN  ALGEBEA 

Adding  r^  and  rg,  we  have 


-  b  +  Vja  _  4  ac  -  J  -  Vj2  _  4  ae 

-26 
2a 

h 

'■i  '  '^-                           2« 

a 

Therefore                  —  (r^  +  ^2)  ~  "  * 

Multiplying  r^  by  r^^  we  have 

-4ac 

4  a2                4  ^2      a 

Therefore                       r^^  =  -  • 

These  results  may  be  expressed  verbally  as  follows: 
For  the  equation  ax^  +  ftjr  +  c  =  0, 

/.   The  sum  of  the  roots  with  its  sign  changed  is  -• 

c  ^ 

II.   The  product  of  the  roots  is  -  • 

These  relations  frequently  afford  the  simplest  means  of 
checking  the  result  of  solvmg  a  quadratic  equation,  as 
illustrated  in  Exercises  18-31,  below. 


EXERCISES 

Form  the  equation  whose  roots  are  the  following : 

solution.     -:(l-i\=-L^^.,  (?\/i\=_5  =  £. 
\2      5/  10      a    \2/\   ;5/  6      a 

Hence  the  required  equation  is 

a;2  -  —  a:  -  5  =  0,  or  10  a;2  -  7  a:  -  12  =  0. 
10         5 


THEORY  OF  QUADRATIC  EQUATIONS        ITl 


2. 

hh 

3. 

h-h 

4. 

H,  -  H- 

5. 

4.41,  1.59. 

6. 

2  +  3V3,  2- 

-3  Vs. 

7. 

4  +  V-  2,  4 

-V-2. 

8. 

f  +  Vt,  1— 

V7. 

9. 

HiV-3, 

i-iV: 

11. 

±V-1 

12. 

1 

a,  —-' 
a 

13. 

S  a    5  a 

2'  2 

1  +  1. 


14.  1  +  a,  1 


a. 


10.  If  ±  V3. 


a  -{-b    a  —  h 

^^'  T'  — TT' 

a  —  0    a  -\-  0 

16.  6,  8,  f 

17.  -  4,  4,  f 


Solve  the  following  equations   by  the  use  of  the  formula 
and  check  the  result  by  the  use  of  I  and  II,  above  r 


18.  cc2_5ic  +  6  =  0. 

19.  x'-x-^^O. 

20.  x2-2x-4  =  0. 

21.  x^- 9a; -10  =  0. 

22.  a;2  +  2a;  +  l  =  0. 

23.  ^2  + 8a: +  16  =  0. 

24.  £c2  +  5a;  +  5  =  0. 

25.  2r»2_|_3x- 6  =  0. 


26.  3cc2+3£c-5  =  0. 

27.  5a;2-6a;  +  10=0. 

28.  x^  4-  £c  +  1  =  0. 
1 


29 


-  +  -  +  - 
3  ^4^5 


0. 


30.  -  +  4x-7  =  0. 

31.  Qx'-^.x- 


0. 


Find  the  value  of  the  literal  coefficient  in  the  following : 

32.  a;^  +  2  i^  —  c  =  0,  if  one  root  is  2. 

33.  0?^  —  a?  —  c  =  0,  if  one  root  is  6. 

34.  a^  —  ca-  —  70  =  0,  if  one  root  is  7. 

35.  ar^  +  2  ^>x  +  20  =  0,  if  one  root  is  -  4. 

36.  a?^  —  8  a;  +  c  =  0,if  one  root  is  twice  the  other. 

37.  a;^  +  7  a:  +  c  =  0,  if  one  root  exceeds  the  other  by  2. 

38.  x^  + 11 X  +  ^  =  0,  if  the  difference  between  the  roots  is  10. 


_^,+V62- 

-  4  ac 

2^ 

-h-^h^- 

-  4  ac 

172  SECOND  COURSE  IN  ALOEBEA 

99.  Character  of  the  roots  of  a  quadratic  equation.    It 

is  often  desirable  to  determine  whether  the  roots  of  a  given 
quadratic  equation  are  real  or  imaginary,  rational  or  irra- 
tional, equal  or  unequal,  without  solvmg  the  equation. 
This  can  be  accomplished  by  use  of  the  formulas  for  the 
roots  of  the  quadratic  ax^  +  5a;  +  c  =  0  : 

(1) 

r,=  ^^^^~^-^.  (2) 

These  expressions  are  seen  to  differ  from  each  other  only 
in  the  sign  preceding  the  radical.    The  expression 

ft2  -  4  ac, 

which  appears  under  the  radical  sign,  is  called  the  dis- 
criminant of  the  quadratic.  The  only  way  in  which  r^ 
or  ^2  can  be  a  complex  number  is  for  the  discriminant 
to  be  negative.  If  all  of  the  coefficients  a,  5,  and  c  are 
rational,  r^  or  r^  can  be  rational  only  when  Vj2  —  ^ac  is 
rational ;  that  is,  when  the  discriminant  is  a  perfect  square. 
Hence  if  a,  6,  and  c  are  rational,  an  inspection  of  (1) 
and  (2)  shows  that  the  following  statements  are  true: 

/.  If  y^—  4iac  is  positive  and  not  a  perfect  square,  the 
roots  are  real,  unequal,  and  irrational. 

For  example,  ina:^  —  8a;  +  2  =  0  the  discriminant  h^  —  4:ac  equals 
(—  8)*  —  4  •  1  •  2  =  56,  which  is  not  a  perfect  square.  The  roots  of  the 
equation  are  the  real,  unequal,  and  irrational  numbers  4  ±  vl4. 

//.  If  y^  —  ^ac  is  positive  and  a  perfect  square,  the  roots 
are  real,  unequal,  and  rational. 

For  example,  in  the  equation  2 a;^  —  3a;  —  9  =  0  the  discriminant 
62  —  4  ar  equals  (—  3)^  —  4-2-(— 9)=  81,  wfiich  is  a  perfect  square. 
The  roots  are  the  real,  unequal,  rational  numbers  —  J  and  +  3. 


THEORY  OF  QUADRATIC  EQUATIONS        173 

///.  i/  ^2  _  4  ^^  {g  2^ro,  the  roots  are  equal. 

For  example,  in  the  equation  4:x^  —  12x  +  9  =  0,  the  discrimi- 
nant P  -  4  ac  equals  (-  12)2  _  4  •  4  •  9  =^  144  -  144  =  0.  The  only 
number  which  satisfies  this  equation  is  f ,  so  in  one  sense  the  equa- 
tion has  only  one  root.  But  since  the  left  member  has  two  identical 
factors  each  of  which  affords  the  same  root  of  the  equation,  it  is 
customary  to  say  that  the  equation  has  equal  roots. 

IV.  If  IP'  —  4:  ac  is  negative^  the  roots  are  imaginary. 
For  example,  in  the  equation  2  x^  —  b  x  -\-  ^  =  0  the  discrimi- 
nant 62  _  4  ac  equals  (-  5)2  -  4  •  2  •  4  =  25  -  32  =  -  7.    The  roots 

of  the  equation   are  the  conjugate   imaginaries and 

+  5-V~7  ^ 

4 

ORAL  EXERCISES 

Determine  the  character  of  the  roots  of  the  following 
equations  without  solving: 

1.  ic'-f  5aj  +  6  =  0.  Q.  x'-{-x+l=0. 

2.  3a;2-4ic+l=0.  7.  x^- 8a; +16  =  0. 

3.  2x2- 6a;- 3  =  0.  8.  2x2h-3x  +  5  =  0. 

4.  2x'-3x-2  =  Q.  9.  4x2-4x+l=0. 

5.  5x2-5x  +  4  =  0.  10.  3^2 -2x- 2  =  0. 


EXERCISES 

Determine  the  value  of  k  which  will  make  the  roots  of  the 
following  equations  equal : 

1.  x2-A;x-M6  =  0. 

Solution.    a  =  l,h=-k,c  =  lQ. 

Hence  '         h'^-^ac  =  k^-  64. 

In  order  for  the  roots  to  be  equal,  U^  —  ^ac  must  equal  zero. 

Therefore  P  -  64  =  0. 

Whence  1-  =  ±  8. 


174  SECOND  COURSE  IN  ALGEBRA 

Check.    Substituting  8  for  k  in  the  original  equation, 

a;2  -  8  X  +  16  =  0. 
Whence  x  =  4,  only. 

Similarly,  substituting  —  8  for  k, 

x^  +  8  X  +  IQ  =  0. 
Whence  x  =  —  4,  only. 

2.  x^-^kx-j-16  =  0.  7.  9cc2  +  30£c  +  A;  4-  9  =  0. 

3.  x^  -  lOcc  -\-k  =  0.  S.  4.kx^-  60ic  +  25  =  0. 

4.  2x^-hSx  +  k  =  0.  9.  9k^x^-S4:X-\-4:9  =  0. 

5.  x^-3kx  +  S6  =  0.  10.  49x2  -  (A:  +  3)a;  +  4  =  0. 

6.  3£c2  +  4A;x  +  12  =  0.  11.  (k"" -\- 5) x^ -  SO x -{- 25  =  0. 

Determine  the  values  of  a  for  which  the  following  systems 
will  have  two  sets  of  equal  roots : 

12.  ^'  ^  ^  "^^  13.  ^'  +  ^'  =  "^'^  14.  ^2/  =  «^' 

'2/  =  ic  +  a.  *y  =  x  +  l.  *a;H-2/  =  l. 

100.  Number  of  roots  of  a  quadratic.  In  section  87  we 
found  that  every  quadratic  equation  has  two  roots. 

Up  to  the  present  we  have  assumed  that  a  quadratic 
equation  can  have  no  other  root  than  the  ones  found  by 
the  method  of  completing  the  square.  This  fact  can  be 
proved  as  follows: 

Proof.   If  we  writ/e  the  equation  ax^  +  62;  +  c  =  0  in  the  form 

x2  ^^,  ^  ^  E  -  0    and    substitute    therein    from    (I)    and   (II)   on 
a  a 

page  170,  we  get  x^  —  (r^  +  r^)  x  +  r^r^  =  0.    This  can  be  factored 

and  written  as  (x  —  r{)  (x  —  r^)  =  0.  Now  if  any  value  of  x  different 

from  rj  and  rg,  say  r,  be  a  root  of  this  equation,  such  a  value  when 

substituted  for  x  must  satisfy  the  equation  (x  —  r^)  (x  —  r^)  =  0. 

Hence  (r  —  r^)(r  —  r^)  must  equal  zero.    By  assumption,  however, 

r  is  different  from  r^  and  r^.    Consequently  neither  the  factor  r  —  r^ 


THEORY  OF  QUADRATIC  EQUATIONS        175 

nor  r  —  r^  can  equal  zero,  and  therefore  their  product  cannot  equal 
zero.  This  proves  that  no  additional  value  r  can  satisfy  the  equation 
x^  —  (r^  +  rg)  a:  +  r^^  —  0.  As  this  equation  is  but  another  form  of 
03?"  +  62;  +  c  =  0,  the  latter  has  only  two  roots. 

101.  Factors  of  quadratic  expressions.    Let  r^  and  r^  be 

the  roots  of  a^  +  5a:  +  c  =  0 ;  then 
I        c 

or  a7^-\-  hx  -\-  c  =  a(x  —  r-[)  (x  —  r^. 

Therefore  the  three  factors  a,  x  —  r-^^  and  x  —  r^  of  any 
quadratic  expression  can  be  found  if  we  first  set  the  ex- 
pression equal  to  zero  (see  section  34)  and  solve  the  equa- 
tion thus  formed.  Obviously  the  character  of  the  roots  so 
obtained  will  determine  the  character  of  the  factors.  Hence 
by  the  use  of  the  discriminant  l^—  4:ac  we  can  decide 
whether  the  factors  of  a  quadratic  expression  are  real  or 
imaginary,  rational  or  irrational,  without  factoring  it. 

EXERCISES 

Determine  which  of  the  following  expressions  have  rational 
factors : 

1.  a;2-3cc-40.        3.  Tic^-Qx  +  lS.       5.  120? -11x^1. 

2.  2a:2  +  5aj-7.       4.  ^^.x'-x-lO.       6.  5ic2  +  3cc-20. 

7.  3^2  -  9a-  +  28.  9.  x"  -  2«cc  +  {a?  -  b^. 

8.  33  A^  -  233  A,  -  6.  10.  ahx"  -  {W-  +  a')x  +  ah. 
Separate  into  rational,  irrational,  or  imaginary  factors : 
11.  2x2  +  5x-8. 

Solution.    Let  2  a:2  +  5  a:  -  8  =  0. 


Solving  by  formula,    x  = — ^^ ^  = 


176  SECOND  COURSE  IN  ALGEBRA 


rr.                         -5+V89       ^           -5-V89 
1  hen  r,  = and  r„  = 


4 
Therefore   2x^-{-5x 


=  2\x 


-S+Vsoir        _5-V89' 
|(4x  +  5-V89)(4a;  +  5  +V89). 


12.  x'-lx-T.  21.  £c2  +  7£f  +  8. 

13.  x'-Ax-  1.  22.  x^-\-x  +  l. 

14.  x^-}-2x-^2.  23.  x^  +  1. 

15.  a^2  4.4^_9.  24.  £c2  +  9. 

16.  4a;2_l2£c-9.  25.  cc^  -  2 ao:  +  a'^  -  ^». 

17.  25x2  +  20ic  +  4.  26.  a:^ +  6acc  +  9'a2_4^ 

18.  6;z^2^14£c-40.  27.  4£c2  +  4aic  +  a^  _  4c. 

19.  10-9a;-9a;2.  28.  x""  -  ^ ax  +  4: a^ -\- c. 

20.  10x2  4-12-26cc.  29.  a^t-^  _,_  ^^  _|_  ^ 

30.  x^  —  xy-j-  5x  —  2y  +  6. 

Solution.   Let       x^  —  xy  •{■  5 x  —  2i/  +6  =  0. 
Then  x^  +  (6  -  y)x  -  2 y  +  Q  =  0. 

Solving  for  x  in  terms  of  y  by  the  formula, 


(5  -  y)  ±  V (5  -  ?/)2  -  4  (-  2  y  +  6) 


_  -5  +  y±-y/y^-2y  +  l 

~  2  '     ' 

Whence  x  =—  2  and  y  —  3. 

Therefore  x^  -  xy  -\- 5  x  -  2  y  -\-  Q  =  (x  -\-  2)(x  -  ?/  +  3). 

31.  Sx^  -  6xy  -\-14:X  -  4:y  +  S. 

32.  x^-xy-2f-\-Sx-6y. 

33.  ar^  -  4xy  -  ?/  +  Sy-  -  2  -  cc. 


CHAPTER  XIV 
GRAPHS  OF  QUADRATIC  EQUATIONS  IN  TWO  VARIABLES 

102.  Graph  of  a  quadratic  equation  in  two   variables. 

Before  solving  graphically  a  quadratic  system,  the  method 
of  graphing  one  quadratic  equation  m  two  variables  must 
be  clearly  understood. 

EXAMPLES 

1.  Construct  the  graph  of  ic^  =  3  y. 

Solution.     Solving  the  equation  for  y,  y  =  — . 

o 

We  now  assign  values  to  x  and  then  compute  the  approximate 
corresponding  values  of  y.    Tabulating  the  results  gives : 


If       x  = 

6 

4 

3 

2 

1 

0 

- 1 

—  2 

-4 

-6 

then  y  = 

12 

^/ 

3"" 

1 

i 

0 

h 

t 

¥- 

12 

Using  an  a;-axis  and  a  y-SLxis 
as  in  graphing  linear  equations, 
plotting  the  points  correspond- 
ing to  the  real  numbers  in  the 
table,  and  drawing  the  curve 
determined  by  these  points,  we 
obtain  the  graph  of  the  adja- 
cent figure.  The  curve  is  called 
a  parabola. 

The  graph  of  any  equation  of 
the  form  x^  =  ay  is  a  parabola. 


Y 

\ 

p 

AR 

AB 

)Li 

^ 

1 

^ 

c 

mi 

IPHQ] 

7 

^ 

X 

•i— 

3y 

Y 

J 

\) 

q 

f 

V 

J 

A 

/ 

X 

r\ 

s 

1 

/ 

X 

-2 

j- 

0 

1 

>     I 

\ 

Y' 

1.77 


178  SECOND  COUESE  IN  ALGEBRA 

2.  Graph,  the  equation  xy  -{-  S  =  0. 

8 
Solution.    Solving  for  y,  y  = 

X 

Assigning  values  to  x  as  indicated  in  the  following  table,  we 
then  compute  the  corresponding  values  of  2^ : 


If      x  = 

-6 

-5 

-4 

-3 

-2 

-1 

-i 

\ 

1 

2 

3 

4 

5 

6 

8 

then  y  — 

1 

f 

2 

1 

4 

8 

16 

-16 

-8 

-4 

-! 

-2 

-1 

-i 

-1 

Proceeding  as  before  with  the  numbers  in  the  table,  we  obtain 
the  two-branched  curve  of  the  figure  below,  which  does  not  touch 
either  axis.    The  curve  is  called  a  hyperbola. 


Y 

y 

Q 

HI 

fPE 

RB 

OL 

A 

A 

Y 

0 

H 

X 

X 

-7 

-^ 

-. 

l-X 

0  ] 

. 

I  3  ^ 

t 

yA 

H 

U^ 

GRAPH  OF 

-2 

X 

^:2/H 

■8- 

=0 

f 

r 

_J 

The  graph  of  any  equation  of  the  form  xy  =  K  is  a 
hyperbola.  The  curve  for  xy  —  K  (Jr=any  constant)  is 
always  in  the  same  general  position ;  that  is,  if  K  is  posi- 
tive, one  branch  of  the  curve  lies  in  the  first  quadrant  and 
the  other  branch  in  the  third.  If  A'  is  negative,  one  branch 
lies  in  the  second  quadrant  and  the  other  in  the  fourth. 


GRAPHS  OF  QUADRATIC  EQUATIONS        179 

3.  Graph  the  equation  x-  -\-  y^  —  X^. 

Solution.    Solving  for  y,  y  =  ±  Vl6  —  x^. 

Assigning  values  to  x  as  indicated,  in  the  following  table,  we 
obtain  from  page  274  the  corresponding  values  of  y : 


If  x  = 

-5 

-4 

-3 

-2 

-1 

0 

1 

2 

3 

4 
0 

5 

theni/  = 

±3V^ 

0 

±2.64 

±3.46 

±3.87 

±4 

±3.87 

±3.46 

±2.64 

±3V^ 

For  values  of  x  numerically  greater  than  4  it  appears  that  y  is 
imaginary.  The  points  corresponding  to  the  pairs  of  real  numbers 
in  the  table  lie  on  the  circle  in  the  accompanying  figure.  The  center 
of  the  circle  is  at  the  origin, 
and  the  radius  is  4. 

The  graph  of  any  equa- 
tion of  the  form  oP-  -\-  tf^^  r^ 
is  a  circle  whose  radius  is  r. 
This  can  be  proved  from 
the  right  triangle  FKO, 
If  P  represents  any  point 
on  the  circle,  OK  equals 
the  2;-distance  of  P,  KF 
equals  the  ^/-distance,  and 
OF  equals  the  radius.  Now  ~0K^ ■\-KF'^  =0F'^ \  that  is, 
o?'-\-y^  —  r^.  It  follows,  then,  that  the  graphs  of  2^-|-?/2=9 
and  a^  +  «/2  =  8  are  circles  whose  centers  are  at  the  origin 
and  whose  radii  are  3  and  V8  respectively.  Hereafter, 
when  it  is  required  to  graph  an  equation  of  the  form 
2^  4-  ?/2  =  7^,  the  student  may  use  compasses  and,  with  the 
origin  as  the  center  and  with  the  proper  radius  (the  square 
root  of  the  constant  term),  describe  the  circle  at  once. 

In  all  the  graphical  work  which  follows,  the  student  will  save 
time  by  obtaining  from  the  table  on  page  274  the  square  roots  or 
cube  roots  which  he  may  need. 


Y 

^A 

M 

' 

Q 

H 

k. 

/ 

r,s 

•■  V 

2 

> 

/ 

\ 

C 

IRC 

LE 

X' 

\ 

X 

K 

- 

0  ] 

,      I 

>      J 

i 

V 

/ 

\ 

k 

-2 

J 

/ 

N 

K 

^ 

l> 

Y 

GRAPH  OJ 

i' 

^■'f2/i  =  16 

r 

180 


SECOND  COURSE  IN  ALGEBRA 


4.  Graph  the  equation  I6x^  -\-9y^  =  144. 

Solution.    Solving,  y  =  ±  ^  V9  —  x^.   Proceed   as   in    Example    3 


If      x  = 

-4 

-3 

-2 

-1 

0 

+  1 

+  2 

+  3 

+  4 

then  y  = 

±^v^ 

0 

±2.98 

±3.77 

'±4 

±3.77 

±2.98 

0 

±^V^ 

For  any  value  of  x  numerically  greater  than  3,  y  is  imaginary. 
The  points  corresponding  to  the  real  numbers  in  the  table  lie  on 
the  graph  of  the  adjacent  figure. 
The  curve  is  called  an  ellipse. 

The  graph  of  any  equation 
of  the  form  of  ax^  -\-by^  =  c 
in  which  a  and  b  are  unequal 
and  of  the  same  sign  as  c  is 
an  ellipse. 

Note.  These  three  curves  — 
the  ellipse,  the  hyperbola,  and  the 
parabola — were  first  studied  by 
ihe  Greeks,  who  proved  that  they 
are  the  sections  which  one  obtains 
by  cutting  a  cone  by  a  plane.  Not 
for  hundreds  of  years  afterwards  did  anyone  imagine  that  these  curves 
actually  appear  in  nature,  for  the  Greeks  regarded  them  merely  as 
geometrical  figures,  and  not  at  all  as  curves  that  have  anything  to 
do  with  our  everyday  life.  One  of  the  most  important  discoveries  of 
astronomy  was  made  by  Kepler  (1571-1630),  who  showed  that  the 
earth  revolves  around  the  sun  in  an  ellipse,  and  stated  the  laws 
which  govern  the  motion.  Those  comets  that  return  to  our  field  of 
vision  periodically  also  have  elliptic  orbits,  while  those  that  appear 
once,  never  to  be  seen  again,  describe  parabolic  or  hyperbolic  paths. 

The  path  of  a  ball  thrown  through  the  air  in  any  direction, 
except  vertically  upward  or  downward,  is  a  parabola.  The  approxi- 
mate parabola  which  a  projectile  actually  describes  depends  on  the 
elevation  of  the  gun  (the  angle  with  the  horizontal),  the  quality  of 
the  powder,  the  amount  of  the  charge,  the  direction  of  the  wind,  and 
various  other  conditions.    This  makes  gunnery  a  complex  subject. 


Y 

y^'      H^ 

Y-            3-         "Si 

^                  3               I 

-:        «>        e 

2                 ^ 

t               V            \ 

t 

tL Lx 

-2     -           0             2 

i.±  m 

,                    "^      ELLIPSE      / 

^                    2                         t 

\             -2                         f- 

\                               J- 

S.            z 

GRAPH  OF   >jv^          ^ 

16xH  91/^14' 

III               Y 

GRAPHS  OF  QUADRATIC  EQUATIONS        181 

EXERCISES 

Construct  the  graphs  of  the  following  equations  and  state 
the  name  of  each  curve  obtained: 

1.  x'  =  4:y.  3,  x'+i/  =  49.         5.  x^  -  f  =  16. 

2.  y^-{-2x  =  0.  4.  x^-{-t/  =  18.         6.  xij  =  12. 

7.  xy  =  -6.  9.  1603^-9?/^  =  144. 

8.  9x''-\-4.y^  =  S6.  10.  25 x^ -\- 9 if  =  225. 

103.  Graphical  solution  of  a  quadratic  system  in  two 
variables.  That  we  may  solve  a  system  of  two  quadratic 
equations  by  a  method  similar  to  that  employed  in  section  44 
for  linear  equations  appears  from  the 


EXAMPLES 


1.   Solve  graphically  [^^^_^^^  _  {j 


(1) 
(2) 


Solution.  Constructing  the  graphs  of  (1)  and  (2),  we  obtain  the 
straight  line  and  the  parabola  shown  in  the  adjacent  figure.  There 
are  two  sets  of  roots  correspond- 
ing to  two  points  of  intersection, 
which  are 

Note.  If  the  straight  line  in 
the  adjacent  figure  were  moved 
to  the  right  in  such  a  way  that 
it  always  remained  parallel  to  its 
present  position,  the  points  A  and 
B  would  approach  each  other  and 
finally  coincide.  The  line  would 
then  be  tangent  to  the  parabola 
at  the  point  x  =  4,  y  =  1. 

Were  the  straight  line  moved  still  farther,  it  would  neither  touch 
nor  intersect  the  parabola  and  there  would  be  no  graphical  solution 
(see  page  182). 


Y 

\ 

^^ 

\ 

A 

^ 

v^ 

\ 

Q 

N 

\ 

o 

N 

.<2) 

\ 

N 

X 

^ 

L 

K 

-^ 

~  1 

> 

o\ 

L     \ 

> 

5 

r 

> 

b\ 

> 

^ 

y 

1) 

^ 

^ 

\ 

^ 

F' 

182 


SECOND  COUESE  IN  ALGEBEA 


2.  Solve  grapMcally  i 


20^  +  2/ =  12,  (1) 

y  +  4:x  =  19.  (2) 

Solution.  The  graphs  of  (1)  and  (2)  are  the  straight  line  and 
the  parabola  of  the  adjacent  figure.  These  curves  have  no  real 
points  of  intersec- 
tion. There  are, 
however,  two  pairs 
of  imaginary  roots. 
Solving  (1)  and 
(2)  by  substitution, 
X  =  JJ-  +V^  or 
^7^-  —  v  —  1,  and 
y  =  l-2V^  or 
1  +  2V-1. 

The  essential 
point  to  be  em- 
phasized here  is 
that  real  roots 
of  a  simultane- 
ous system  cor- 
respond to  real  intersections,  and  imaginary  roots  correspond 
to  no  intersections  of  real  graphs. 


ex 
3.  Solve  graphically  | 


x^-f  =  4, 

£c  -  6  =  0. 


(1) 
(2) 

Solution.  Constructing  the  graphs  of  (1)  and  (2),  we  obtain  the 
hyperbola  and  the  parabola  of  the  figure  on  the  opposite  page. 
There  are  four  sets  of  roots  corresponding  to  the  four  points  of 
intersection,  which  are  approximately 


=  -2.7. 


1^  =  3.1.      ^1^  =  1.8.  ^ 


a:  =-2.7, 
y=-1.8. 


3.7, 
-3.1. 


If  the  two  curves  had  been  so  chosen  as  to  intersect  only  twice, 
their  equations  would  have  had  only  two  sets  of  real  roots. 


GRAPHS  OF  QUADRATIC  EQUATIOKS        188 

Examples  1,  2,  and  3  partially  illustrate  the  truth  of  the 
following  statement: 

If  in  a  system  of  two  equations  in  two  variables  one 
equation  is  of  the  mth  degree  and  one  .of  the  nth,  there 
are  usually  mn  sets  of  roots  (real  or  imaginary)  and  never 
more  than  mn  such  sets. 


r 

\ 

/^ 

\ 

^^'^ 

A 

>■ 

/ 

\ 

s. 

^^ 

Y_ 

m 

— 

. 

\ 

^£ 

/ 

l/ 

^ 

\ 

/ 

X 

' 

y' 

f 

X 

V 

-^ 

0    ] 

\  ? 

^ 

J 

\ 

/ 

^ 

* — 

-2 

\ 

V 

) 

i!l 

/ 

/ 

-\ 

^ 

— ■ 

■— 

/ 

-4 

\( 

1) 

y 

Y' 

EXERCISES 

If  possible,  solve  graphically  each  of  the  following  systems 


2/^  =  4  cc, 

x"  4-  tf  =  25, 
2.  -^ 


2/ 


3. 


2x  =  10. 


4. 


5. 


x2/=  6, 


10. 


11. 


x^2y 


=  17, 
9. 


^2^4y  =  17, 
x-}-22/  =  12. 


«2  +  7/2  =  25, 

x^  +  2/^  =  9, 
a;  4-  ?/  =  10. 
0^2  +  ^2  =  36, 
x^-y'  =  25. 

12 


13 


a^2  +  y^=9, 

^-   X^  -  7/2  =  16. 

x?/  =  12, 
^'  2a; +  2/ =  10. 

0^2  =  4  7/, 

x'-\-y  =  h, 
if  +  x  =  Z. 
ic_2/V8  =  0,     . 
x^  =  7/8  -  9  7/. 


9. 


184 


SECOND  COURSE  IN  ALGEBRA 


104.  Graphical  presentation  of  numerical  data.  A  great 
variety  of  statistics  can  be  presented  graphically  in  a  very 
striking  manner.  Business  and  commercial  houses  have 
during  the  past  few  years  used  the  method  extensively 
not  only  to  present  facts  but  also  to  aid  in  interpreting 
them  and  in  indicating  their  tendencies. 

The  following  exercises  illustrate  some  of  the  possibili- 
ties in  the  graphical  presentation  of  numerical  data. 


EXAMPLE 

For  the  year  1912  the  total  income  and  expenses  .per  mile 
of  line  of  all  the  railroads  in  the  United  States  having  a 
yearly  revenue  of  one  million  dollars  or  more  was  as  follows : 


Jan. 

Feb. 

Mar. 

Apr. 

May 

June 

July 

Aug. 

Sept 

Oct. 

Nov. 

Dec. 

Income  in  dollars  . 
Expenses  in  dollars 

930 
730 

970 
710 

1050 
750 

980 
720 

1040 
740 

1080 
750 

1120 

760 

1220 

780 

1210 

780 

1320 
830 

1220 
810 

1170 
800 

■ 

Delia 

rs 

> 



y 

y' 

V. 

B 

Nl« 

^ 

"»^ 

^ 

V 

^ 

uu 

^ 

— - 

^ 

-^ 

_ 







-1 



_ 







i-i 

PE 

NS 

E- 

__ 

_. 

=» 



— 

— 

— 

- 

— 

— 

- 

— 

— 

— 

— 

— 

— 

— 

— 

OO 

RC 

2 

T 

- 

- 

=- 

•— 

V 

^ 

"^ 

^ 

•ja 

n. 

F< 

b. 

M 

ar. 

A 

Dr. 

M 

ay 

Ju 

n. 

Ji 

1. 

Ai 

Iff. 

s« 

p. 

0 

k. 

N 

w. 

D 

^c. 

- 

GRAPHS  OF  QUADRATIC  EQUATIONS        185 

The  foregoing  graph  represents  the  given  data  on  the  same 
axes. 

The  lowest  curve  on  the  diagram  shows  the  profits  of  the 
railroads.  It  was  obtained  by  plotting  the  differences  of  the 
numbers  in  the  table. 

EXERCISES 

1.  The  average  incomes  of  155  members  of  a  certain  college 
class  for  the  first  ten  years  after  their  graduation  is  given  in 
the  following  table : 


Years  after  graduation 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

Income  in  dollars  .   .   . 

706 

902 

1199 

1651 

2039 

2408 

2382 

2709 

3222 

3804 

Graph  the  data.    What  tendencies  do  you  note  '! 

2.  The  number  of  inches  of  rainfall  during  the  month  of 
July  and  the  number  of  bushels  of  corn  yielded  per  acre  for  a 
term  of  years  in  a  certain  locality  is  given  in  the  following 
table : 


Year 

'89 

'90 

'91 

'92 

•  '93 

'94 

'95 

'96 

'97 

'98 

'99 

'00 

'01 

'02 

Rainfall 

Corn  yield  .... 

5.4 
32 

2.6 

23 

5.1 

27 

3.7 

27 

3.4 
24 

1.9 

18 

4.8    5.4 
30     38 

3.7 
25 

4.3 

26 

4.6 
28 

4.7 
30 

1.2 
19 

6.0 
32 

Plot  these  data  on  the  same  axes..  How  do  you  account  for 
the  similarity  of  the  curves  ? 

3.  The  numbers  of  hundreds  of  telephone  calls  in  certain 
business  and  residential  sections  of  New  York  City  are  given 
in  the  following  table  for  various  hours  of  the  day.  Plot  both 
sets  of  data  on  the  same  axes  and  explain  the  reason  for  differ- 
ences in  the  shapes  of  the  graphs. 


Time  of  day 

7 

8 

■9 

10 

11 

Noon 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

Business  district  . 
Residence  district 

1 
.25 

2 

4 
35 

55 
73 

108 
76 

101 
65 

78 
52 

75 

48 

93 

47 

87 
40 

67 
38 

35 
38 

8 
36 

2 
32 

1.5 
25 

1 
15 

1 
4 

1 
3 

RE 


186 


SECOND  COURSE  IN  ALGEBRA 


4.   Measurements  of  the  breadth  of  the  heads  of  a  thousand 
students  in  a  certain  school  were  as  follows : 


Head  breadth  in  inches 
Number  of  students  .  . 

,5.5 
3 

5.6 
12 

5.7 
43 

6S 

80 

5.9 
131 

6.0 

236 

6.1 

185 

6,2 
142 

6.3 

99 

6.4 
37 

6.5 

15 

6.6 

12 

6.7 
3 

6.8 
2 

Construct  a  graph  of  the  above  data. 

5.  The  chest  measurements  of  10,000  soldiers  were  tabulated 
as  follows : 


Chest  measure-  -» 

ment  in  inches  / 

Number  of  soldiers 


322 


38 
1305 


39 

18G7 


40 

1882 


41 
1628 


42 
1148 


38 


48 


Construct  a  graph  of  the  above  data. 

It  may  appear  accidental  that  the  foregoing  measure- 
ments should  group  themselves  with  any  regularity.  But  if 
the  number  of  measurements  of  this  type  is  large  and  each 
is  made  with  care,  they  obey  a  law  called  the  law  of  proba- 
bility. In  fact  the  graph  of  the  data  in  Exercises  4  and  5  is 
a  close  approximation  to  what  is  called  the  probability  curve. 

The  equation  of  this  curve  is  y  =  e"^*  when  eJ  =  2.7 
approximately. 

6.  Construct  the  graph  of  the  equation  y  =  e'^"^  between 
the  values  cc  =  —  2  and  x  =  1. 


Hint.   Let  x  =  —  2,  —  |, 


\,  0,  etc. 


Note.  It  is  well  established  that  physical  characteristics,  such 
as  those  illustrated  by  the  graphs  of  Exercises  4  and  5,  obey  the 
law  of  probability.  If  the  graph  of  Exercise  4  is  carefully  consid- 
ered, it  may  raise  the  question,  Do  mental  characteristics  also  obey 
the  law?  An  interesting  aspect  of  this  is  given  by  the  fact  that  an 
increasing  number  of  high  schools  and  colleges  assume  that  such 
is  the  case,  and  grade,  or  mark,  their  students  by  a  system  based 


GEAPHS  OF  QUADRATIC  EQUATIONS        187 

on  the  law  of  probability.  Such  a  system  assumes  that  if  one  hun- 
dred or  more  students  in  any  subject  are  examined,  the  number  of 
students  and  their  degree  of  mastery  of  the  subject  arrange  them- 
selves according  to  the  probability  "curve  shown  below.  Obviously 
between  very  many  students  the  differences  in  grades  attained  will 
be  small,  between  many  others  they  will  be  moderate,  and  between 
only  a  few  will  they  be  great. 


In  the  statistical  study  of  problems  which  have  their  origin  quite 
remote  from  each  other,  this  curve  frequently  occurs,  and  occupies  a 
central 'position  in  the  mathematical  theory  of  statistics. 


CHAPTER  XV 

SYSTEMS   SOLVABLE  BY  QUADRATICS 

105.  Introduction.  The  general  equation  of  the  second 
degree  in  two  variables  is  ax^  +  hy^  +  cxy  -\-dx  +  ey  -i-f  =  0. 
To  solve  a  pair  of  such  equations  requires  the  solution  of 
an  equation  of  the  fourth  degree.  Even  the  solution  of 
x^-{-  y  =  3  and  ^2  _j_  ^.  _  3  requires  the  solution  of  such  an 
equation.  In  fact  only  a  limited  number  of  systems  of  the 
second  degree  in  two  variables  is  solvable  by  quadratics. 
By  the  graphical  methods  of  Chapter  XIV  the  student  can 
solve  graphically  for  real  roots  any  system  of  quadratic 
equations,  provided  the  terms  have  numerical  coefficients. 
The  algebraic  solution  of  such  systems  will  in  many  cases 
be  possible  for  him  only  after  further  study  of  algebra. 

106.  Linear  and  quadratic  systems.  Every  system  of 
equations  in  two  variables  in  which  one  equation  is  linear 
and  the  other  quadratic  can  be  solved  by  the  method  of 
substitution. 

EXAMPLE 


Solve  the  svstem  i         ,   *  *    ^  „  ' 
-^           la;  +  2  7/  =  13. 

(1) 
(2) 

Solution.    Solving  (2)  for  y  in  terms  of  x, 

1^-x 
y       2      ' 
Substituting    — - —  for  y  in  (1), 

(8) 

.^-..f«r)=r. 

(4) 

From  (4),               2  x"  _  13  x  -  7  =  0. 

188 

(5) 

SYSTEMS  SOLVABLE  BY  QUADRATICS       189 

Solving  (5),  a:  =  7  or  —  ^. 

Substituting  7  for  a:  in  (3), 

13-7 

=  3. 
Substituting  —  ^  iov  x  in  (3), 


y-- 

13  + Jj 
2 
=  6i 

The  two  sets  of  roots 

^vQx  =  l,y-- 

=  3,  and  ; 

r  =- 

-i,y  =  6|. 

Check.   Substituting 

7  for  X  and  3  for  y  in 

(1) 

and  (2), 

49  -  42  = 

=  7, 

« 

7+6  = 

=  13. 

Substituting  -  i  for 

X  and  6f  for 

3/  in  (1) 
=  7, 
=  13. 

and 

(2), 

EXERCISES 

Solve  the  following  systems,  pair  results,  and  check  each 
set  of  roots : 


1. 

x-\-y  =  5, 

2. 

2x-^y  =  l. 

3. 

x^  -\-xy  =  15, 
x  +  y  =  ^. 

4. 

x''  +  2xy  =  21, 
x  +  2y  =  ^. 

5. 

x'  +  2y  =  ll, 
2x-y  =  2. 

A 

^_10:r  =  6. 

x4-3y  =  13. 


8 

2  ar2-a;y  =  70, 

o» 

ic+-4y  =  23. 

Q 

^2  + 25?^  =  108, 

V  • 

4s-3^=6. 

10. 

5^-352^  =  22, 

4s  +  2^  =  2. 

11. 

5^  +.  ^2  ^  169, 

2s  +  ?J  =  22. 

12. 

s^  +  2  ^=^  =  27, 
4^-25  =  18. 

0 

SECOND  COURSE  IN  ALGEBRA 

13. 

s^  +  Sts-\-t"  =  44, 
2s-t  =  0. 

22.i  +  |  =  a, 

14. 

s^-\-ts  +  t^  =  12, 
s-\-t  =  2. 

1      X      5 
•       2/      2~4* 

15. 
16. 

2s^-ts  -{-t^  =  16, 
2s-t  =  5. 

6      4_2 

s       t~  S' 

4:X  -\-  y  =  6. 

o.    ^  +  13 

4.^  +  ^^=28,  6-  +  1      2 

•  2a:2  +  3a^2/-98.  ^  +  4_1 


-4-^  =  2, 


2/       .T       6 


19.  -4-^  =  2,  32/  +  2x  =  2. 

?/      a? 

3^5 +  22/ =  5.  26    ^'  + '^'  +  2^^  =  40, 

*  s-\-t +  2  =  0. 
on    ^       1       1 


Sx-y  =  2. 


y  —  X  =  V2, 


X-     2,  98    ^'  +  y'  +  4x  +  6./  =  40, 

10_10^3  29^  +  ^^^  =  ^^ 

y        X       2  •  ^Z^  +  x^zzzieo^. 

107.  Homogeneous  equations.  If  both  the  equations  of 
a  system  are  quadratic,  an  attempt  to  solve  it  by  substi- 
tution usually  gives  an  equation  of  the  fourth  degree.  In 
most  cases  such  an  equation  could  not  be  solved  by  fac- 
toring, and  at  the  present  time  its  solution  by  any  other 
method  is   beyond    the    student.     With   certain   types   of 


■  SYSTEMS  SOLVABLE  BY  QUADRATICS       191 

systems,  however,  which  occur  more  or  less  frequently,  we 
can  employ  special  devices  and  avoid  the  solution  of  any 
equation  of  higher  degree  than  a  quadratic.  Among  these 
systems  are  the  so-called  "  homogeneous  "  systems. 

An  equation  is  homogeneous  if,  on  being  written  so  that 
one  member  is  zero,  the  terms  m  the  other  member  are  of 
tHe  same  degree  with  respect  to  the  variables. 

Thus  x^  +  y^  =  xy  and  x^  —  oxy-{-y^  =  0  are  homogeneous  equa- 
tions of  the  second  degree.  2x^  -{•  y^  =  x'^y  —  3  xy"^  is  a  homogeneous 
equation  of  the  third  degree. 

The  system  \     {     ^  „       '  )►  is  a  homogeneous  system. 

Systems  like  i  ^    „        „      ,        '       ^  are  often  called  homogeneous 
1,2  a:^  —  2/2  4-  4  a;y  =  6  J 

systems,  but  strictly  speaking  they  are  not.  As  will  be  seen,  the 
method  of  solving  such  systems  is  about  the  same  as  the  method 
of  solving  a  homogeneous  system.  Hence  they  are  classed  with 
homogeneous  systems. 

108.  Systems  having  both  equations  quadratic.  Occasion- 
ally when  both  equations  are  quadratic  the  terms  which 
occur  in  the  two  are  so  related  that  the  elimination  of 
terms  involving  both  variables  or  of  the  constant  terms 
can  be  performed.    The  following  system  is  of  such  type: 


EXAMPLES 


x" 


1.  Solve  the  system  r;    ,  ^2/ "  2,  (1) 

Solution.    First  eliminate  xy  by  addition  (§  47), 

(1)  -4,  4  a:2  -  4  a:y  =  8.  (3) 

(2) +  (3),  7x2  =  28. 

Whence  j:  =  ±  2. 


192 


SECOND  COURSE  IN  ALGEBRA 


Substituting  +  2  for  x  in  (1),   4  —  2  ?/  =  2,   whence   y  =  \. 
Substituting  —  2  for  x  in  (1),   4  +  2  ?/  =  2,   whence    ^/  =  —  1. 
Therefore  a:  =  2,  -  2, 

and  '  ?/  =  1,  —  1. 

These  values  may  be  checked  as  usual. 

The  method  of  solving  a  system  in  which  every  term  in  each 
equation  except  the  constant  terms  is  of  the  second  degree  is 
as  follows : 

-xy-\-'iif  =  ^,  (1) 

x^-\-y'  =  10.  (2) 


2.   Solve 


Solution.    First  we  combine  the  two  equations  to  obtain  a  homo- 
geneous equation : 

(1)  -5,  5  xy  +  15  2/2  :==  30. 

(2)  -3,  3  a:2  +  3  2/2  =  30. 

(3)  -  (4),  -  3  x2  +  5  xy  +  12  y2  =  0. 


Solving  (5)  for  x  in  terms  of  y, 
and 

Substituting  ^y  iox  x  in  (2), 
From  (8), 

Substituting  from  (9)  in  (6), 
4 


x^^y, 
4« 

9  ?/2  +   ^,2  ^  10.- 

ar=±3. 


16  7/2 


Substituting  -  —^  for  a;  in  (2),     — -^  +  y'^ 


From  (10), 

Substituting  from  (11)  in  (7), 


10. 
y=±^Vi0. 


(3) 
(4) 
(5) 
(6) 

(7) 
(8) 
(9) 

(10) 
(11) 


When        X  = 

3 

-3 

+  tVio 

-^VIo 

then           2/  = 

1 

-  1 

-^  VlO 

+  '^Vio" 

Each  pair  of  values  can  be  checked  as  usual. 


SYSTEMS  SOLVABLE  BY  QUADRATICS       193 

A  quadratic  system  in  which  one  equation  is  homoge- 
neous is  easier  to  solve  than  the  system  of  Example  2,  as 
can  be  seen  from  what  follows: 

3.  Solve  the  system  |^.  _^  ^  _  5  ^  ^  3^  ^2) 

Hints.    Solving  (1)  for  x,         x  =  2  </,  ,  (3) 

and  ^  =  \-  (^) 

o 

We  may  now  substitute  from  (3)  in  (2)  and  from  (4)  in  (2),  solve  the 
resulting  equations,  and  then  complete  the  solution  as  in  Example  2. 

EXERCISES 

Solve,  pair  results,  and  check  each  set  of  real  roots : 

Zi/-\-x'  =  2d>.  2x^-^xy  +  if=(). 

'^3.^  +  2.^  =  7,  2a^-.^  +  2^  =  12, 

2x  -^xy  =  -^.  8.  2x''+xy-{-2f=^. 

^if-xy  =  2, 
^*  2 2/=^ +  3 0^1/ =  38.  9    20^2  + 2/2  =  3a;y, 

x^  +  ^xy  =  lQ. 
y"^  —  xy  =  2  x^^ 

'  2x^-{-xy  =  16.  _    52_3^^^4 


5    x''-Sy-{-S  =  0, 


''']^-,y=,r 


y^-\-x'-25.  s^  +  2st-\-4.t^  =  lS, 

6.  Solve  Example  3  completely.         "  t^  +  S  —  —  2st 

109.  Symmetric  systems.  A  system  of  equations  in 
X  and  1/  is  symmetric  if  the  system  is  not  altered  by  sub- 
stituting X  for  y  and  1/  for  x. 

Thus  x^  +  ^2  _  g  and  x  +  y  =  11  is  a  symmetric  system,  but 
x^  —  y^  =  Q  and  z  -{-  y  ~  11  is  not. 


194  SECOND  COURSE  IN  ALGEBRA 

Certain  symmetric  systems  or  systems  which  are  nearly 
symmetric  can  be  easily  solved  by  the  method  of  substi- 
tution.   Of  such  the  following  are  types: 

1x1/ =  4:.  \x±y=2.  l2a;±«/  =  ll. 

A  few  other  systems  which  are  symmetric  or  nearly  so 
are  more  easily  solved  by  certain  special  methods.  The 
following  list  contains  typical  systems,  and  the  methods 
applicable  are  given  in  Exercises  1,  10,  and  12. 


EXERCISES 

1.  Solve  <^      ^^^  )^{ 

\x2j  =  6.  (2) 

Hints.    These  equations  can  be  combined  in  such  a  way  as  to  obtain 
definite  values  ior  x  +  y  and  x  —  ?/  as  follows  : 

(2).  2,                                                2xy  =  12.  (3) 

{l)  +  (3),                          a;2  + 2x^  +  2/2  =  49.  (4) 

From  (4),                                        x-\-  y=  ±7.  (5) 

(l)-(3),                         x^-2xy  +  y^  =  26.  (6) 

From  (6),                                        x-y  =  ±6.  .    (7) 
(5)  and  (7)  combined  give  four  systems  of  equations  : 

rx  +  2/  =  7,                 (8)                        fx-{.y=-7,  (11) 

-^\x-y  =  6.                (9)                       \x-y  =  5.  (9) 

rx+7/  =  7,                 (8)                        fx^y=-7,  (11) 

^\x-2/=-6.           (10)                      \x-?/=-5.  (10) 

The  solution  of  systems  A^  B,  C,  and  D  is  left  to  the  student. 

The  pairs  of  roots  found  for  the  four  systems  A,  B,  C,  and 
D  will  check  in  the  original  system. 

.r*^  +  ;//=:100,  x''  +  xy  +  9f==19,     ^    x^-\-4y^  =  20, 

xy  =  48.  *  xy  =  6.  '  o-y  =  4. 


SYSTEMS  SOLVABLE  BY  QUADRATICS       195 


9x'  +  t/  =  ei, 

xy  =  10. 

4.x''-{-xy  +  f  =  So, 
xy  =  12. 

9. 

6 
x  =  -■ 

y 

7 
"3 

4x^  +  252/^  =  41, 
'•  xy  =  2. 

10. 

x'      y' 

1  =  6. 

xy 

x'-xy-\-y'  =  S, 

Hint.  To  clear  of  fractions  in  Exercise  10  would  merely  increase  the 
difl&culty  of  solution.  Instead  we  solve  in  a  manner  similar  to  that  of 
Exercise  1  (see  Exercise  6,  p.  78). 

1  =  12.  ■ 


xy 

Then 

1 

X2 

+ 

^  +  i  = 

xy      j/2 

:25. 

Whence 

1 

X 

+  '  = 

y 

:±5. 

In  like  manner 

1 

X 

1_ 

±1. 

—  =  12.     . 
xy 

13. 

11        13 

?>x^  2y~   6' 

11         97 

9ic2  '  4y^"36* 

11       5 

~1  +  ~5  =  7' 
^^    x^      ^r      4 
12. 

1       1  _  1 

x      y~  2 

14. 

x^      x^^2/^       '' 

i  =  6. 

^2/ 

XI            12       1 
Hints.    — 1 = 

x2       xy      ?/2 

Then                       —  = 
xy 

1 

"4" 
=  1, 

etc 

15. 

x^      xy      f 
l  +  i  =  7. 

196  SECOND  COUKSE  IK  ALGEBRA 

110.  Equivalent  systems.    Equivalent  systems  of  equations 
are  systems  which  have  tlie  same  set  or  sets  of  roots. 
If  the  two  systems 

r^    ,^  =  12,      (1)    ^^^       r.    ,  =  2,       (8) 

lx  +  y  =  ^.  (2)  Vx-\-y  =  Q.  (1) 

are  solved,  only  one  pair  of  roots,  x  —  4:  and  y  =  2,  is  obtained 
for  A  and  the  same  pair  for  B.  Systems  A  and  B  are 
equivalent,  though  usually  a  system  which  consists  of  a 
linear  and  a  quadratic  has  two  sets  of  roots  and  hence 
cannot  be  equivalent  to  a  linear  system  (see  page  183, 
second  paragraph). 

Sometimes  an  equation  simpler  than  either  of  those  given 
in  a  system  can  be  derived  from  this  system  by  dividing 
the  left  and  right  members  of  the  first  equation  by  the  corre- 
sponding members  of  the  second.  In  systems  like  those  of 
the  following  Ust  the  equation  so  obtained  taken  with  one  of 
the  first  two  gives  an  equivalent  system  more  easily  solved 
than  the  original  one. 

EXERCISES 

Solve,  using  division  where  possible,  pair  results,  and  check 
each  set  of  real  roots  : 


""  x  +  2y^2. 

4. 

4a^^-y  =  16, 

2x  +  y  =  S. 

Hint.    Division  gives  the 

equiv- 

5. 

R^h  -  75  =  0, 

alent  system 

Rh  =  15. 

x-22/  =  10, 
X  —  y  =  1. 

6. 

X      y 

^-  2y-x=-\l. 

7. 

x'-¥y'  =  2^, 

a;  +  2/  =  4. 

SYSTEMS  SOLVABLE  BY  QUADRATICS       197 


In  the  following  figure,  (1),  (2),  and  (3)  are  the  graphs  of 
x^  -{-  y^  =  28,  x^  —  xy  •\-  i/  =  7,  and  x  -\-  y  =  ^  respectively. 
These  equations  are  all  used  in  the  solution  of  Exercise  7.  The 
graph  makes  clear  in  a  striking  way  that  the  system  (1)  and  (3) 
is  equivalent  to  the  system  (2)  and  (3). 


JF                           1 

^ 

3)^^ 

S^ 

\ 

X    4^       ^ 

1^               ^ 

"^^^       =^  S 

~^    "^--^^^ 

^'^.>      l^^S 

-7        ^     J-^-A 

/       1          "^ 

/                     >\ 

X' ,:    t  ^^ :^ 

/-2     -1          U      1        2y      3       \ 

^       4-       1      i^           ^ 

-yr              -1      ^1          ^ 

^T                 ^        \        '3:^ 

4^^             y"          \          % 

^---^                 ^ 

(1^. 

J5 

\ 

Y 

a^«-8y^  =  35, 
x  —  2y  =  h. 


X       y 


1, 


y"^      x^ 


X       y 


152, 


MISCELLANEOUS  EXERCISES 

Solve  by  any  method  and  pair  results.  If  any  system  can- 
not be  solved  algebraically  by  the  methods  previously  given, 
solve  it  graphically. 


a!  4-  2/  =  4. 
•  x-'dy  =  \, 


'    £C  +  «/  =  7. 


198  SECOIS'D  COURSE  IN  ALGEBRA 

^^o      A  i«    0^^  + 2^2/ +  22/^  =  10, 

•  5.  y  "^  '  3x'-xi/-f  =  51. 

Q    (x  +  yf  =  9,  '  x'  +  y  =  11. 

•  (^_3/).  =  49.  ^^    ^.  +  ^,^  +  ^.  =  7, 


^•.^ 


2aj2_^2/2^33,  x2  +  2/'  =  10. 

-^  072  4-0-?/  + 2/ =  0, 

3  ^2  _  8  A)2  =  40,  *  £c2  +  ici/  +  a;  =  0. 
5  ^2  +  ^^2  ^  81 

—  4-  —  =  13 

4i2f  +  3  =  9i2|,  -                    0)-^^/ 

12i2f  +  i2|-^9^.  1       1 


=  1. 

^^'  a;^  +  y  =  20.  -^8-^=^ 

0^2  +  2/2  =  169,  23.  ^      ^'^ 

a?       ?/~    * 


xy  -\-  X  =  18, 


ajy  =  60, 


24^  40.2  +  2/2  =  289, 
£C2/  =  60. 

25    0^^  +  2/^  =  9, 
'  a; V  =  20. 

•  x  +  2/  =  5- 
27.  9x  -^2/=lS  =  ^2/- 

28.4.  +  J  =  46  =  ?|^-i. 

5  Tr,2- 6.8  >F|  =  99.55, 

TF2-Tr|  =  20.  _i_  +  — !— =  -, 

20    ^-2      2/-2      4 

0^2^  +  22/^  =  2,  l_l  =  i. 

^  '   3iC2^  +  62/^  =  2,  X      y      12 


12. 

x^  =  y, 
xy  =  S. 

13 

x  —  xy  =  5, 

AO. 

2y-\-xy  =  6. 

14. 

x'-y'  =  19, 

x'-\-xy-\-7/  =  19. 

4  7^2  +  7  ?/i2  =  9, 

15. 

2  7^2  _    9   ^  ^,,2^ 

SYSTEMS  SOLVABLE  BY  QUADEATICS        199 
88,  x-1 

34. 


30. 

X  —  2/  =  6. 

31. 

xy  =  c, 

X  -\-  y  =  a. 

32. 

x-^  +  y-^  =  2. 

33. 

3, 


35. 
36. 
37. 
PROBLEMS 


2/-1 

/  +  y+l_13 

x^-x+1      43' 

4a;^-13a;y+9V=9, 

^¥  —  y^  =  ^' 

x^-\-2xy=16, 
2>x'-4.xy-\-2f=  6. 

a-«  =  ^  +  37, 
x^y  z^xy^  -\- 12. 


(Reject  all  results  which  do  not  satisfy  the  conditions  of  the  problems.) 

1.  Find  two  numbers  whose  difference  is  6  and  the  differ- 
ence of  whose  squares  is  120. 

2.  The  sum  of  two  numbers  is  20  and  the  sum  of  their 
squares  is  202.    Find  the  numbers. 

3.  Find  two  numbers  whose  sum  plus  their  product  is  132 
and  whose  quotient  is  3. 

4.  It  takes  k)^  rods  of  fence  to  inclose  a  rectangular  lot 
whose  area  is  one  acre.    Find  the  dimensions  of  the  lot. 

5.  The  area  of  a  right  triangle  is  180  square  feet  and  its 
hypotenuse  is  41  feet.    Find  the  legs. 

6.  The  area  of  a  pasture  containing  15  acres  is  doubled  by 
increasing  its  length  and  its  breadth  by  20  rods.  What  were 
the  dimensions  at  first  ? 

7.  The  difference  of  the  areas  of  two  squares  is  495  square 
feet  and  the  difference  of  their  perimeters  is  60  feet.  Find 
a  side  of  each  square. 

8.  The  area  of  a  rectangular  field  is  43^  acres  and  one 
diagonal  is  120  rods.    Find  the  perimeter  of  the  field. 


200  SECOND  COUESE  IN  ALGEBRA 

9.  The  value  of  a  certain  fraction  is  |-.  If  the  fraction  is 
squared  and  44  is  subtracted  from  both  the  numerator  and  the 
denominator  of  this  result,  the  value  of  the  fraction  thus 
formed  is  -^^.    Find  the  original  fraction. 

10.  Two  men  together  can  do  a  piece  of  work  in  4|-  days. 
One  man  requires  4  days  less  than  the  other  to  do  the  work 
alone.    Find  the  number  of  days  each  requires  alone. 

11.  The  perimeter  of  a  rectangle  is  250  feet  and  its  area 
is  214  square  yards.    Find  the  length  and  the  width. 

12.  The  base  of  a  triangle  is  8  inches  longer  than  its  alti- 
tude and  the  area  is  1^  square  feet.  Find  the  base  and  altitude 
of  the  triangle. 

13.  The  volumes  of  two  cubes  differ  by  316  cubic  inches 
and  their  edges  differ  by  4  inches.    Find  the  edge  of  each. 

14.  The  perimeter  of  a  right  triangle  is  80  feet  and  its  area 
is  240  square  feet.    Find  the  legs  and  the  hypotenuse. 

15.  The  perimeter  of  a  rectangle  is  7  a  and  its  area  is  a^. 
Find  its  dimensions. 

16.  A  man  travels  from  A  to  B,  30  miles,  by  boat  and  from 
B  to  C,  120  miles,  by  rail.  The  trip  required  6  hours.  He 
returned  from  C  to  B  by  a  train  running  10  miles  per  hour 
faster,  and  from  B  to  A  by  the  same  boat.  The  return  trip 
took  5  hours.    Find  the  rate  of  the  boat  and  of  each  train. 

17.  There  were  1400  fewer  reserved  seats  at  a  certain  sale 
than  of  unreserved  seats,  and  the  price  of  the  latter  was 
15  cents  less  than  the  price  of  the  former.  The  total  proceeds 
were  |490,  of  which  |250  came  from  the  reserved  seats.  Find 
the  number  of  each  kind  of  seats  and  the  price  of  each. 

18.  If  a  two-digit  number  be  multiplied  by  the  sum  of  its 
digits,  the  product  is  324;  and  if  three  times  the  sum  of 
its  digits  be  added  to  the  number,  the  result  is  expressed  by 
the  digits  in  reverse  order.   Find  the  number. 


SYSTEMS  SOLVABLE  BY  QUADEATICS       201 

19.  The  sum  of  the  radii  of  two  circles  is  31  inches  and  the 
difference  of  their  areas  is  155  tt  square  inches.   Find  the  radii. 

20.  The  yearly  interest  on  a  certain  sum  of  money  is  $42. 
If  the  sum  were  |200  more  and  the  interest  1%  less,  the 
annual  income  would  be  |6  more.  Find  the  principal  and 
the  rate. 

21.  A  vheelman  leaves  A  and  travels  north.  At  the  same 
time  a  second  wheelman  leaves  a  point  3  miles  east  of  A  and 
travels  east.  One  and  one-third  hours  after  starting,  the  short- 
est distance  between  them  is  17  miles,  and  3|-  hours  later  the 
distance  is  53  miles.    Find  the  rate  of  each. 

22.  A  starts  out  from  P  to  Q  at  the  same  time  B  leaves  Q 
for  P.  When  they  meet,  A  has  gone  40  miles  more  than  B. 
A  then  finishes  the  journey  to  Q  in  2  hours  and  B  the  journey 
to  P  in  8  hours.  Find  the  rates  of  A  and  B  and  the  distance 
from  P  to  Q. 

23.  A  leaves  P  going  to  Q  at  the  same  time  that  B  leaves 
Q  on  his  way  to  P.  From  the  time  the  two  meet,  it  requires 
6|  hours  for  A  to  reach  Q,  and  15  hours  for  B  to  reach  P. 
Find  the  rate  of  each,  if  the  distance  from  P  to  Q  is  300  miles. 


GEOMETRICAL  PROBLEMS 

1.  The  sides  of  a  triangle  are  6,  8,  and  10.    Find  the 
altitude  on  the  side  10. 

Hint.    From  the  accompanying  fig- 
ure we  easily  obtain  the  system 

(x^  +  y^  =  S6, 
\  (10  -  x)2  +  ?/2  =  64. 

2.  The  sides  of  a  triangle  are 
8,  15,  and  17.  Find  the  altitude 
of  the  triangle  on  the  side  17  and  the  area  of  the  triangle. 

3.  The  sides  of  a  triangle  are  11,  13,  and  20.    Find  the 
altitude  on  the  side  20  and  the  area  of  the  triangle. 

UE 


202 


SECOND  COURSE  IK  ALGEBRA 


4.  The  sides  of  a  triangle  are  13,  14,  and  15.  Find  the 
altitude  on  the  side  14  and  the  area  of  the  triangle. 

5.  The  sides  of  a  triangle  are  12,  17,  and  25.  Find  the 
altitude  on  the  side  12  and  the  area  of  the  triangle. 

6.  Find  correct  to  two  decimals  the  altitude  on  the  side  16 
of  a  triangle  whose  sides  are  16,  20,  and  24  respectively. 

7.  The  parallel  sides  of  a  trapezoid  are  14  and  26  respec- 
tively, and  the  two  nonparallel  sides  are  10  each.  Find  the 
altitude  of  the  trapezoid. 

Hint.   Let  ABCD  be  a  trapezoid.   Draw  CE  parallel  to  DA  and  CF 
perpendicular  to  AB. 
Then 


EC  =  10, 


and 


AE  =  14, 


and 


EB  =  26-  14,  or  12. 
If  we  let  EF  =  x,  FB  must  equal  12  —  x ;  then  we  can  obtain  the 


system  of  equations 


rx2  +  ?/  ::::  IQO, 
\  (12  -  X)2  +  2/2  =  100. 


8.  The  two  nonparallel  sides  of  a  trapezoid  are  10  and  17 
respectively,  and  the  two  bases  are  9  and  30  respectively.  Find 
the  altitude  of  the  trapezoid. 

9.  The  bases  of  a  trapezoid  are  15  and  20  respectively, 
and  the  two  nonparallel  sides  are  29  and  30.  Find  the  alti- 
tude of  the  trapezoid  and  the  area. 

10.  The  sides  of  a  trapezoid  are  7,  9,  20,  and  24.  The  sides 
24  and  9  are  the  bases.   Find  the  altitude  and  the  area. 

11.  The  sides  of  a  trapezoid  are  21,  27,  40,  and  30.  The 
sides  21  and  40  are  parallel.  Find  the  altitude  and  the  area 
of  the  trapezoid. 


SYSTEMS  SOLVABLE  BY  QUADEATICS       203 


12.  The  sides  of  a  trapezoid  are  23,  85,  100,  and  x.  The 
sides  23  and  100  are  the  bases,  and  each  is  perpendicular  to 
the  side  x.  Find  x 
and  the  area  of  the 
trapezoid. 

13.  The  area  of  a 
triangle  is  1  square 
foot.  The  altitude 
on  the  first  side  is 
16  inches.  The  sec- 
ond side  is  14  inches 
longer  than  the  third. 

Find  the  three  sides.         

L  B 

14.  A  rectangular 

tank  is  8  feet  6  inches  long  .and  6  feet  8  inches  wide.  A  board 
10  inches  wide  is  laid  diagonally  on  the  floor.  What  two  equa- 
tions must  be  solved  to  determine  the  length  of  the  longest 
board  that  can  be  thus  laid  ? 


HiKTS.    Let  BR 
the  triangle  AKL. 


X  and  DE  =  y.   The  triangle  DKE  is  similar  to 


CHAPTER  XVI 
PROGRESSIONS 

111.  A  sequence  of  numbers.  In  all  fields  of  mathe- 
matics we  frequently  encounter  groups  of  three  or  more 
numbers,  selected  according  to  some  law  and  arranged  in 
a  definite  order,  whose  relations  to  each  other  and  to  other 
numbers  we  wish  to  study. 

There  is  an  unlimited  variety  of  such  groups,  or  suc- 
cessions, of  numbers.  Only  two  simple  types  will  be 
considered  here. 

112.  Arithmetical  progression.    An  arithmetical  progression 

is  a  succession  of  terms  in  which  each  term  after  the  first  is 
formed  by  adding  the  same  number  to  the  preceding  one. 

Thus,  if  a  denotes  the  first  term  and  d  the  common  number  added, 
any  arithmetical  progression  is  represented  by 

a,  a  -h  d,  a  -{•  2  d,  a  +  ^  d,  a  +  4:  d,  '  •  ■ . 

This  common  number  d  is  called  the  common  difference 
and  may  be  any  number,  positive  or  7ieijative,  It  may  be 
found  for  any  given  arithmetical  progression  by  subtract- 
ing any  term  from  the  term  which  follows  it. 

The  numbers  3,  7,  11,  15,  •  •  •  form  an  arithmetical  progression, 
Muce  any  term,  after  the  first,  minus  the  preceding  one  gives  4.  Simi- 
larly, 12,  6,  0,  —  6,  — 12,  •  •  •  is  an  arithmetical  progression,  since 
any  term,  after  the  first,  minus  the  preceding  one  gives  the  com- 
mon difference  —  6.  In  like  manner  5,  5,  6^,  •  •  •  is  an  arithmetical 
progression  whose  common  difference  is  1^. 

204 


PKOGEESSIONS  205 

ORAL  EXERCISES 

State  the  first  four  terms  of  the  arithmetical  progression  if : 

1.  a  =  2,d  =  3.  6.  a  =  100,  d  =  -10. 

2.  a  =  5,  d  =  4..  7.  a  =  20x,  d  =  —  2x. 

3.  a  =  10,  d=  6.  S.  a  =x,  d  =  2x. 

^.  a  =  20,  d  =  5.  9.  a  =  17  m,  d  =  —  2  m. 

5.   a  =  18,  ^--3.  10.  a=V5,d  =  l-\-V5. 

From  the  following  select  the  arithmetical  progressions,  and 
in  each  arithmetical  progression  find  the  common  difference : 

11.1,10,19,....  16.  8,9f,10|,.... 

12.  4,  12,36,....  17.  15,3,  -12,.-.. 

13.  19,  11,  3,  .  .  ..  18.  6a,  10a,  14«,  •  •  .. 

14.  9,  12,  1,  16,  .  .  ..'  19.  18  a,  14 a,  12  a,  •  .  .. 

15.  3,  ^,  -V-,  ....  20.  5  a,  5  a  +  3,  5  a  +  6,  .  .  .. 

21.  Sx,  X  —  2,  —X  —  4:,  •  .  .. 

22.  2x-l,  X,  1,  2-cc,  ...."^  ^' 

23.  —  5  Va,  —  2  Va,  Va,  4  Va,  .... 

24.  5V^-1,  4V^-2,  3V^-3,  .... 

113.  The  last  or  Trth  term  of  an  arithmetical  progression. 
In  the  arithmetical  progression 

a,  a  +  d,  a  +  2  d,  a  +  3  (f,  a  +  4  ff,  •  .  . 

one  observes  that  the  coefficient  of  d  in  each  term  is  1 

less  than   the   number   of   the  term.    Hence   the  ?ith,   or 

general,  term  is  a -\- (n  —  1)  d.    li  I  denotes  the  nth  term, 

we  have 

l=a+(n-i)d. 


206  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 

Eind   the   required    terms    of    the    following    arithmetical 
progressions : 

1.  The  fifteenth  term  of  2,  7,  12,  •  •  •. 
Solution.    Here  a  =  2,  c?  =  5,  and  n  =  15. 
Hence  I  =  a  -{■  (n —lyd 

becomes  Z  =  2  +  14  •  5  =  72. 

2.  The  eighth  term  of  1,  4,  7,  •  •  •. 

3.  The  eleventh  term  of  15,  9,  3,  -  3,  •  •  .. 

4.  The  twenty -first  term  of  a,  4  a,  7  <7,  •  •  •. 

5.  The  sixteenth  term  of  Qcc,  jr,  —  7ic,  •  •  •. 

6.  The  sixth  term  of  ^^-,  f ,  |,  •  •  •• 

7.  The  eleventh  term  of  —  y,  |-,  1,  •  •  • . 

8.  The  sixth  term  of  V3,  4  V3,  7  Vs,  •  •  •. 

9.  The  eighth  term  of  3  Vs,  V5,  -  V5,  -  3  Vs,  .... 

10.  The  eighth  term  of  5  +  a,  7+  3 a,  9  +  5 «,  .  .  -. 

11.  The  twelfth  and  the  twentieth  terms  of  0,  —  5x  +  2, 
4 -10a;,  .... 

12.  The  tenth  term  of  V^  +  3,  5  Va  +7,  9  Va  +  11,  .  .  .. 

13.  The  tenth  and  the  twentieth  terms  of  9  Va  —  3,  5  Va, 

Va  4- 3,  .... 

■yj^     5  "v  £C     9  "v  £C 

14.  The  fourteenth  term  of  -—>  — - — >  — r— 5  '  '  '• 

z       z        z 

15.  The  ninth  and  the  twelfth  terms  of  V3,  — 7=?  3  Vs,  . .  ..' 

V3 

V  iC  3   V  cc 

16.  The  eighth  and  the  sixteenth  terms  of  -jr-  4- 1,  —75 h  2, 


5  Vcc 


+  3, 


17.  The  fifteenth  term  of  ^^  -  12,  -  5,  ^^^    ^     ""^ 


PROGRESSIONS  207 

18.  The  twenty-ninth  term  of  a,  a  -\-  d,  a  -\-  2  d,  -  -  -. 

19.  The  mth  term  of  a,  a  -^  d,  a  -\-  2  d,  •  -  • . 

20.  The  (m  —  l)th  term  ot  a,  a  -\-  d,  a  -{-  2d,  •  •  •. 

21.  The  (n  -  2)th  term  oi  a,  a -\- 5,  a -{-10,  ■  ". 

o  _ 

22.  The  (n  -  5)th  term  of  —^,  3  V3,  5  Vs,  •  •  -. 

23.  rindthe(7i-3)thtermofV5-l,2V5-2,3(V5-l),  .... 

24.  Find  the  nth  term  of  ->  ?  2  —  -?  •  •  •• 

a        a  a 

25.  The  first  and  third  terms  of  an  arithmetical  progression 

are  2  and  22.    Find  the  seventh  term ;  the  nth  term. 

26.  The  first  and  second  terms  of  an  arithmetical  progres- 
sion are  r  and  s  respectively.  Find  the  third  term  and  the 
nth.  term. 

27.  The  edges  of  a  box  are  consecutive  even  integers  with 
n  the  least.  Express  in  terms  of  n,  (a)  the  sum  of  the  edges ; 
(b)  the  area  of  the  faces  ;  (c)  the  volume. 

28.  A  body  falls  16  feet  the  first  second,  48  feet  the  next, 
80  feet  the  next,  and  so  on.  How  far  does  it  fall  during  the 
twelfth  second  ?  during  the  nth  second  ? 

29.  The  digits  of  a  certain  three-digit  number  are  in  arith- 
metical progression.  If  their  sum  is  24  and  the  sum  of  their 
squares  is  194,  find  the  number. 

114.  Arithmetical  means.  The  arithmetical  means  between 
two  numbers  are  numbers  which  form,  with  the  two 
given  ones  as  the  first  and  the  last  term,  an  arithmetical 
progression. 

The  insertion  of  one  or  more  arithmetical  means  between 
two  numbers  is  performed  as  in  the  following  examples: 


208  SECOND  COURSE  IN  ALGEBEA 

EXABIPLES 

1.  Insert  four  arithmetical  means  between  7  and  72. 
Solution.    1=  a  +  (n  —  l)d. 

There  will  be  six  terms  in  all. 

Therefore  72  =  7  +  (6  ~  1)  d. 

Solving,  ^  =  13. 

The  required  arithmetical  progression  is  7,  20,  33,  46,  59,  72. 

2.  Insert  one  arithmetical  mean  between  h  and  k. 
Solution.    I  =  a  -\-  (n  —  1)  d. 

There  will  be  three  terms  in  all. 
Therefore  k  =  h  +  2d. 

Solving,  d  =  —^' 

Therefore  the  progression  is  h,  h  -\ — ,  k,  or  7i,  — -— ,  k. 

It  follows  from  the  above  that  the  arithmetical  mean  between  two 
numbers  is  one  half  of  their  sum. 

EXERCISES 
Insert  one  arithmetical  mean  between : 
1.  3  and  15.       2.  3  and  27.      3.  h  and  5  h.      4.  m  and  n. 

Insert  two  arithmetical  means  between  : 

5.  2  and  8.  7.  49  and  217.  9.  x  and  3  //. 

6.  5  and  29.  8.  x  and  x  -{-  6  y.  10.  h  and  k. 

Insert  three  arithmetical  means  between : 

11.  10  and  34.  13.  -  12a;  and  Ux. 

12.-5  and  31.  14.  h  and  k. 

15.  Insert  seven  arithmetical  means  between  —13  and  131. 

16.  Insert  five  arithmetical  means  between  —  |  and  ^^. 


PROGRESSIONS  209 

17.  Insert  four  arithmetical  means  between  —  2  Vs  and 
18  V5. 

18.  Insert  five  arithmetical  means  between  7  a  — Sb  and 

13  a +  9&. 

/-  27 

19.  Insert  six  arithmetical  means  between  VS  and p=- 

2V3 

3 

20.  Insert   two   arithmetical   means   between   7=   and 

4V3-6.  1-^ 

21.  In  going  a  distance  of  1  mile  an  engine  increased  its 
speed  uniformly  from  15  miles  per  hour  to  25  miles  per  hour. 
What  was  the  average  velocity  in  miles  per  hour  during  that 
time  ?    How  long  did  it  require  to  run  the  mile  ? 

22.  At  the  beginning  of  the  third  second  the  velocity  of  a 
falling  body  is  64  feet  per  second,  and  at  the  end  of  the  third 
second  it  is  96  feet  per  second.  What  is  its  average  velocity 
in  feet  per  second  during  the  third  second  ?  How  many  feet 
does  it  fall  during  the  third  second  ? 

23.  The  velocity  of  a  body  falling  from  rest  increases  uni- 
formly and  is  32  feet  per  second  at  the  end  of  the  first  second. 
What  is  the  average  velocity  in  feet  per  second  during  the 
first  second  ?  How  many  feet  does  the  body  fall  during  the 
first  second  ?    the  second  second  ? 

24.  Find  the  average  length  of  twenty  lines  whose  lengths 
in  inches  are  the  first  twenty  odd  numbers. 

25.  Find  the  average  length  of  fifteen  lines  whose  lengths  in 
inches  are  given  by  the  consecutive  even  numbers  beginning 
with  58. 

26.  With  the  conditions  of  Problem  23  determine  the  average 
velocity  per  second  of  a  body  which  has  fallen  for  12  seconds. 

27.  A  certain  distance  is  separated  into  ten  intervals,  the 
lengths  of  which  are  in  arithmetical  progression.  If  the  shortest 
interval  is  1  inch  and  the  longest  37  inches,  find  the  others. 


210  SECOND  COUKSE  IN  ALGEBEA 

115.  Sum  of  a  series.    The  indicated  sum  of  several 
terms  of  an  arithmetical  progression  is  called  an   arith- 
metical series.    The  formula  for  the  sum  of  n  terms  of 
an  arithmetical  series  may  be  obtained  as  follows: 
S  =  a+{a  +  d^+(a  +  ^d)+  .  .  .  J^(l-2d)-^(l-d)+l.     (1) 

Reversing  the  order  of  the  terms  in  the  second  member 
of  (1), 

S  =  l+(l-d)+(l-2d^+  .  .  .+(a  +  2c?)4-(a  +  (^)-h«.      (2) 
Adding  (1)  and  (2), 

2^  =  (a  +  Z)  +  (a  +  0  +  (^  +  0+---+(^  +  0  +  («  +  0 
+  (a  +  Z)  =  n(6)^  +  Z). 

Therefore  S= -(«  +  /). 

EXAMPLE 

Required  the  sum  of  the  integers  from  7  to  92,  inclusive. 
Solution,    n  =  86,  a  =  7,  Z  =  92. 

Substituting  in  5  =  -  (a  +  Z), 

Therefore  the  sum  of  the  integers  from  7  to  92  is  4257. 

'  ORAL  EXERCISES 
Using  the  formula  S  =  -(a-\-l)j  find  the  sum  of  the  following : 

1.  The  six-term  series  in  which  a  =  2  and  I  =17. 

2.  The  ten-term  series  in  which  a  =1  and  I  =  46. 

3.  The  twelve-term  series  in  which  a  =  —  12  and  I  =  21. 

4.  The  seven-term  series  in  which  a  =  S  and  I  =  63. 

5.  The  twenty-term  series  in  which  a  =  5  and  I  =  86. 

6.  The  twelve-term  series  in  which  a  =  —  175  and  I  =125. 


progeessio:n^s  211 


Yt 

The  formula  S  =  -(a  +  l)  enables  one  to  find  the  sum  of 

a  series  if  the  first  term,  the  last  term,  and  the  number  of 
the  terms  are  given.  If,  however,  the  last  term  is  not  given, 
but  instead  the  common  difference  is  given  or  evident,  we 

can  use  a  formula  obtained  by  substituting  m.  S  =  -(a -\- V) 

the  value  of  I  from  the  formula  l  =  a-\-{n  —  1)  c?,  on  page  205. 

VI 

Then  s  =  j.\a -^  a -^  (n -V)d\ 

or  S  =  -[2a  +  (n-l)tf]. 

JL 

EXAMPLE 

Required  the  sum  of  the  first  fifty-nine  terms  of  the  progression 
2,9,  16,-.. . 

Solution,    n  =  59,  a  =  2,  and  d  =  l. 
Substituting  in  5  =  ^  [2  a  +  (/i  -  1)  rf], 

5-  =  ^  (4  +  58  •  7)  =  ^  (410)  =  12,095. 


EXERCISES 
Find  the  sum  of  the  following : 

1.  The  first  eight  terms  of  the  series  in  which  a  =  2  and  c?  =  4. 

2.  The  first  ten  terms  of  the  series  in  which  a.  =  8  and  d  =  5. 

3.  The  first  nine  terms  of  the  series  in  which  a  =  1  and 
d  =  ll. 

4.  The  first  -fifteen  terms  of  the  series   in  which  a  =  17 
and  d  =  S. 

5.  The  first  twenty  terms  of  the  series  in  which  a  =  100  and 
d=-5. 


212  SECOND  COURSE  IN  ALGEBRA 

6.  The  first  eight  terms  of  the  series  2  +  4  4-  6  +  •  •  • . 

7.  The  first  ten  terms  of  the  series  1  +  9  -h  17  +  •  •  •. 

8.  The  first  ten  terms  of  the  series  —  8  +  (—  4)  +  0  +  •  •  • . 

9.  The  first  eighteen  terms  of  the  series  1  +  5  +  9  +  •  •  •. 

10.  The  first  twenty  terms  of  the  series  10  +  8  +  6  -f  •  •  -. 

11.  The  first  twelve  terms  of  1  +  |  +  2,  •  •  .. 

12.  The  first  twelve  terms  of  15  +  12i  +  10  H . 

13.  The  first  one  hundred  integers. 

14.  The  first  one  hundred  even  numbers. 

15.  The  first  one  hundred  odd  numbers. 

16.  Show  that  the  sum  of  the  first  n  even  numbers  is  n{n  + 1). 

17.  Show  that  the  sum  of  the  first  n  odd  numbers  is  n^. 

18.  Find  the  sum  of  the  even  numbers  between  247  and  539. 

19.  How  many  of  the  positive  integers  beginning  with  1 
are  required  to  make  their  sum  861  ? 

Hint.   Substitute  in  the  formula  *S  =  -  [2  a  +  (n  —  1)  d]  and  solve  for  n. 

20.  How  many  terms  must  constitute  the  series  7  +  10+13 
+  ...  in  order  that  its  sum  may  be  242  ? 

21.  Beginning  with  90  in  the  progression  78,  80,  82,  •  •  ., 
how  many  terms  are  required  to  give  a  sum  of  372  ? 

22.  The  second  term  of  an  arithmetical  progression  is  —  2 
and  the  eighth  term  is  22.    Find  the  eleventh  term. 

23.  Find  the  sum  of  the  first  t  terras  of  -> > -^  •  •  •• 

a        a  a 

24.  If  Z  =  25,  a  =  1,  and  <^  =  4,  find  n  and  s. 

25.  If  a  =  -  20,  6^  =  11,  and  s  =  216,  find  n  and  I. 

26.  It  d  =  —  9,  n  =  15,  and  s  =  0,  find  a  and  I. 


PEOGKESSIOiS^S  213 

27.  The  first  and  second  terms  of  an  arithmetical  progres- 
sion are  h  and  k  respectively.    Find  the  sum  of  n  terms. 

28.  If  5  =  9  A,  a  =  12  -  10  h,  and  ti  =  9,  find  I  and  d. 

29.  If  s  =  66  V3,  a  =  -  4  V3,  and  ^Z  =  2  Vs,  find  n  and  I. 

30.  A  clock  strikes  the  hours  but  not  the  half  hours.  How 
many  times  does  it  strike  in  a  day  ? 

31.  A  car  running  15  miles  an  hour  is  started  up  an  incline, 
which  decreases  its  velocity  ^  of  a  foot  per  second.  («)  In  how 
many  seconds  will  it  stop  ?  (J))  How  far  will  it  go  up  the 
incline  ? 

32.  A  car  starts  down  a  grade  and  moves  3  inches  the  first 
second,  11  inches  the  second  second,  19  inches  the  third  second, 
and  so  on.  (a)  How  fast  does  it  move  in  feet  per  second  at  the 
end  of  the  thirtieth  second  ?  (h)  How  far  has  it  moved  in  the 
thirty  seconds  ? 

33.  An  elastic  ball  falls  from  a  height  of  24  inches.  On 
each  rebound  it  comes  to  a  point  \  inch  below  the  height 
reached  the  time  before.  How  often  will  it  drop  before  coming 
to  rest  ?  Find  the  total  distance  through  which  it  has  moved. 

34.  The  digits  of  a  three-digit  nmnber  are  in  arithmetical 
progression.  The  first  digit  is  3  and  the  number  is  20^  times 
the  sum  of  its  digits.    Find  the  number. 

35.  A  clerk  received  $60  a  month  for  the  first  year  and  a 
yearly  increase  of  |75  for  the  next  nine  years.  Find  his  salary 
for  the  tenth  year  and  the  total  amount  received. 

36.  If  a  man  saves  |200  a  year  and  at  the  end  of  each 
year  places  this  sum  at  simple  interest  at  6^,  what  will  be 
the  amount  of  his  savings  at  the  time  of  the  sixth  annual 
deposit  ? 

37.  Assuming  that  a  ball  is  not  retarded  by  the  air,  deter- 
mine the  number  of  seconds  it  will  take  to  reach  the  ground  if 


214  SECOND  COURSE  IN  ALGEBEA 

dropped  from  the  top  of  the  Washington  Monument,  which  is 
555  feet  high.   With  what  velocity  will  it  strike  the  ground  ? 
Hint.   See  Exercise  28,  p.  207. 

38.  A  ball  thrown  vertically  upward  rose  to  a  height  of 
256  feet.  In  how  many  seconds  did  it  begin  to  fall  ?  With 
what  velocity  was  it  thrown  ? 

39.  By  Exercise  28,  p.  207,  it  is  seen  that  a  falling  body  obeys 
the  law  of  an  arithmetical  progression.    Show  from  the  data  of 

that  exercise  that  the  general  formula  S  =  -(2  a  -\-  (n  —  1)  d) 

at'' 
becomes  the  special  one  S  =  ^y  which  is  used  in  physics  for 

all  such  problems. 

40.  A  ball  thrown  vertically  upward  returned  to  the  ground 
6  seconds  later.  How  high  did  it  rise  ?  With  what  velocity 
was  it  thrown  ? 

41.  A  and  B  start  from  the  same  place  at  the  same  time  and 
travel  in  the  same  direction.  A  travels  20  miles  per  hour.  B 
goes  30  miles  the  first  hour,  26  miles  the  second,  22  miles  the 
third,  and  so  on.    When  are  they  together  ? 

Note.  In  the  earliest  mathematical  work  known  a  problem  is 
found  which  involves  the  idea  of  an  arithmetical  progression.  In 
the  papyrus  of  the  Egyptian  priest  Ahmes,  who  lived  nearly  two 
thousand  years  before  Christ,  we  read  in  essence,  "  Divide  40  loaves 
among  5  persons  so  that  the  numbers  of  loaves  that  they  receive  shall 
form  an  arithmetical  progression,  and  so  that  the  two  who  receive 
the  least  bread  shall  together  have  one  seventh  as  much  as  the 
others."  From  that  time  to  this,  the  subject  has  been  a  favorite  one 
with  mathematical  writers,  and  has  been  extended  so  widely  that  it 
would  require  many  volumes  to  record  all  of  the  discoveries  regard- 
ing the  various  kinds  of  series. 

116.  Geometrical  progression.  A  geometrical  progression  is 
a  succession  of  terms  in  which  each  term  after  the  first 
is  formed  by  multiplying  the  preceding  one  always  by  the 
same  number. 


PEOGRESSIONS  215 

Thus,  if  a  denotes  the  first  term  and  r  the  common  multiplier,  then 
any  geometrical  progression  is  represented  by  a,  ar,  ar^,  ar^,  ar\  •  •  • . 

The  common  multiplier  is  called  the  ratio.  It  is  evident 
from  the  above  that  the  ratio  r  in  a  geometrical  progression 
is  found  by  dividing  any  term  by  the  preceding  one. 

The  numbers  2,  10,  50,  250,  •  •  •  form  a  geometrical  progression, 
since  any  term,  after  the  first,  divided  by  the  preceding  one  gives  the 
same  number  5.  Similarly,  the  numbers  3,-3  V2,  6,-6  V2,  •  •  • 
form  a  geometrical  progression,  since  any  term,  after  the  first, 
divided  by  the  preceding  one  gives  the  common  ratio  —  V2. 


ORAL  EXERCISES 

Determine  which  of  the  following  are  geometrical  progres- 
sions and  state  the  ratio  in  each  case : 

1.  1,  3,  9,  27,  ....  7.  i  Ve,  VO,  V54,  .... 

2.  2,4,16,....  8.  Vf,  -V|,2,.... 
3    2,  6,  18,....                          9.  7a,S5a,  175a,.... 

4.  5,  1,  i,  ••••  10.  8V5,  -2V5,  V5,  .... 

5.  18,  -  3,  i  .  .  ..  U.  5x%  lOa^^  20a:^  •  •  .. 

6.  2,  1  V2,  i  V2,  .  .  ..  12.  3y\  12y,  ^Sy\  .  .  .. 

13.  Find  the  condition  under  which  a,  b,  and  c  form  a 
geometrical  progression. 

State   in   order   the   first   four   terms    of   the    geometrical 
progression  in  which  the  first  term  is 

14.  1  and  the  ratio  is  4.  17.  64  and  the  ratio  is  J. 

15.  3  and  the  ratio  is  10.         18.  —  243  and  the  ratio  is  ^. 

16.  —  3  and  the  ratio  is  2.       19.  2  and  the  ratio  is  V3. 


216  SECOND  COURSE  IN  ALGEBRA 

117.  The  nth  term  of  a  geometrical  progression.     Since 
by  definition  any  geometrical  progression  is  represented  by 

a,  ar,  af",  ar\  -  -  -, 

it  is  evident  that  the  exponent  of  r  in  any  term  is  1  Iqss 
than  the  number  of  the  term.  Therefore,  if  t„  denotes  the 
nth,  or  general,  term  of  any  geometrical  progression, 

EXERCISES 

1.  Find  the  fifth  term  of  4, 12,  36,  •  •  -. 

Solution.    Here  a  =  4,  r  =  3,  n  —  1  =  4. 
Substituting  these  values  in  the  formula  ty^  =  ar^^-'^, 

<^  =  4.3*  =  324. 

2.  Find  the  eighth  term  of  1,  2,  4,  •  • .. 

3.  Find  the  tenth  term  of  3,  6,  12,  •  •  -. 

4.  Find  the  eighth  term  of  3,  2,  |-,  •  •  •. 

5.  Find  the  twelfth  term  of  7,  -  14,  28,  •  •  -. 

6.  Find  ^^  of  15,  -5,  +|,  .... 

27  a^ 

7.  Find  t^  of  12  a",  9 a^  -^,  ••-. 

—  2  c*  —  5 

8.  Find  t^  of  — g— ,  -  1,  Y^»  .  .  •• 

9.  Find  fg  of  4  V2,  4,  2  V2, .  •  -. 
1       1    V2 


10.  Find  t„  of 


:>  t:' 


^'       2V2   2    -2 
11.  Write  the  twentieth  term  of  $4.12(1.01),  $4.12(1.01)^, 
H.12(1.01)«,  .... 


PROGRESSIONS  217 

12.  Since  the  nth.  term  of  a  geometrical  progression  is  ar'^~^, 
what  is  the  (n  -  l)th  term  ?  the  (n  -  2)th  ?  the  (n  -  3)th  ?  the 
(n  +  l)th  ?  the  (n  +  2)th  ? 

13.  The  first  and  second  terms  of  a  geometrical  progression 
are  m  and  ti  respectively.    Find  the  next  two  terms. 

118.  Geometrical  means.  Geometrical  means  between  two 
numbers  are  numbers  which  form,  with  the  two  given  ones 
as  the  first  and  the  last  term,  a  geometrical  progression. 

Thus  ar  and  ar^  are  the  geometrical  means  in  a,  ar,  ar^,  a?-^. 

EXAMPLES 

1.  Insert  two  real  geometrical  means  between  9  and  72. 

Solution.  There  are  four  terms  in  the  geometrical  progression, 
and  a  =  9,  n  =  4,  and  tn  =  t^=  72. 

Substituting  these  values  in  t^  =  ar*^-^, 

72  =  9  r3. 

Whence  r  =  2,  and  —  1  ±  V—  3. 

The  required  geometrical  progression  is  9,  18,  36,  72. 

2.  Insert  one  geometrical  mean  between  h  and  k. 

Solution.  There  are  three  terms  in  the  progression,  and  a  =  //, 
w  =  3,  and  t„  =  k. 

Substituting  these  values  in  tn  =  ar"-i,  we  have 

h  —  Ji-r^. 
Solving,  r=±A/-- 

Hence  the  progression  is  ^,  ±  ^-y/-,  ^,  or  A,  ±  VA^,  k. 

It  follows  from  the  above  that  the  geometrical  mean  between 
two  numbers  is  the  square  root  of  their  product. 

K£ 


218  SECOND  COUESE  IN  ALGEBKA 

ORAL  EXERCISES 

Insert  one  geometrical  mean  between : 

1.  1  and  a^.  3.  1  and  4ic^  5.  a  and  a*. 

2.  1  and  x^.  4.3  and  75.  6.  a^  and  d^. 

Insert  two  geometrical  means  between  : 

7.  1  and  icl  10.  1  and  125.  13.  x"-  and  x^. 

8.  1  and  2«.  *    11.  a  and  a\  14.  2  and  16. 

9.  1  and  27.  12.  x  and  x\  15.  5  and  40. 

EXERCISES 

Obtain  progressions  involving  real  terms  only  : 

1.  Insert  two  geometrical  means  between  21  and  168. 

2.  Insert  two  geometrical  means  between  15  and  405. 

3.  Insert  three  geometrical  means  between  3  and  243. 

4.  Insert  one  geometrical  mean  between  9  and  81. 

5.  Insert  one  geometrical  mean  between  aP'  and  a*. 

6.  Insert  three  geometrical  means  between  —  9  and  —  144. 

7.  The  fourth  term  of  a  geometrical  progression  is  16,  the 
eighth  term  is  256.    Find  the  tenth  term. 

8.  The  second  term  of  a  geometrical  progression  is  4  Vs, 
the  fifth  term  is  ^.    Find  the  first  term  and  the  ratio. 

9.  Show  that  the  geometrical  means  between  a  and«c  are 
±  Vac.     ■ 

10.  The  first  and  fourth  terms  of  a  geometrical  progression 
are  a  and  c  respectively.    Find  the  second  and  thii-d  terms. 

11.  Insert  three  geometrical  means  between  h  and  k. 

12.  The  sum  of  the  first  and  fourth  terms  of  a  geometrical 
progression  is  56.    The  second  term  is  6.    Find  the  four  terms. 


PROGRESSIONS 


219 


B       D 
10,  find  BD  and  DC. 


13.  In  the  accompanying  figure,  ABC  is  a  right  triangle 
and  AD  i?,  perpendicular  to  tl^e  hypotenuse  BC.  Under  these 
conditions  the  length  oi  AD  is  always  a  geometrical  mean 
between  the  lengths  of  BD  and  DC. 

{a)  If     BD  =  4:      and    DC  =  9, 
find  AD. 

(b)  If    BD  =  S    and    BC  =  21, 
find  AD. 

(c)  If  BC  =  25  and  AD 

14.  In  the  accompanying  fig- 
ure, AB  touches  and  AD  cuts  the 
circle.  Under  such  conditions 
the  length  of  AB  is  always  a 
geometrical  mean  between  the 
lengths  of  AC  and  AD. 

(a)  If  AD  =  36  and  ^C  =  4, 
-find  AB. 

(b)  If  DC  =  90  and  AB  =  24,  find  AC  and  AD. 

119.  Geometrical  series.  The  indicated  sum  of  n  terms 
of  a  geometrical  progression  is  called  a  geometrical  series.  The 
process  of  obtaining  in  its  simplest  form  the. expression  for 
this  sum  is  often  called  finding  the  sum  of  the  series. 

The  expression  for  the  sum  is  derived  as  follows: 

Let  Sji  denote  the  indicated  sum. 
S  =  a-\-  ar  :{-  ar^  -h  • 


ar' 


-\-ar"-'-\-ar" 


=  a(l-\-r  +  r^  + 
1 


_^^«-3_^^r^-2^^n-l^^^ 


S„  =  a 


d 


(1) 
(2) 


1-r 

since  the  polynomial  in  (1)  is  equal  to  the  fraction  in  the 
parenthesis  of  (2)  (see  Exercise  18,  p.  39). 

a  —  ar" 


Hence 


s„  = 


1-r 


220  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 

1.  Find  the  suni  of  the  first  nine  terms  of  3,  —  6,  12,  •  •  •• 

„  ,   , .                                   ,^        a  —  ar^ 
Solution.  Sn  = 

1  —  r 

By  the  conditions,  a  =  3,  r  =—  2,  and  n  =  9. 


Substituting, 


__3-_H-^ 


l-(-2) 

2.  Find  the  sum  of  1,  5,  25,  •  •  •  to  nine  terms. 

3.  Find  S^for  2,-4,  8,  •  •  •. 

4.  Find  S^  for  40,  20,  10,  ••.. 

5.  Find  ^g  for  -  180,  90,  -  45, .  •  •. 

6.  Find  5^  for  I,  1,1-,.... 

7.  Find  S^^  for  a%  a\  a\.". 

8.  Fina  S^  for  2  Vs,  6,  6  Vs,  •  •  -. 

9.  Find  >S^  for  4,  12,  36,  •  •  .. 

10.  Find  5„  for  125,  -  25  Vs,  25,  .... 

11.  Find  >s;,_i  for  3,  12,  48,  .  .  •. 

12.  Find  S,,  for  3,  -  15,  75,  .... 

13.  Find  »S„,_2  for  x,  4:x\  16x',-  -  -. 

14.  Show  that  for  a  geometrical  progression  S,i  =  — ;-• 

Hint.  Substitute  I  for  the  factor  rtr"-i  in  the  last  term  of  the 
luimerator  in  the  formula  of  Exercise  1. 

15.  A  rubber  ball  falls  from  a  height  of  60  inches  and  on 
each  rebound  rises  60%  of  the  previous  height.  How  far  does 
it  fall  on  its  sixth  descent?  Through  what  distance  has  it 
moved  at  the  end  of  the  sixth  descent  ? 

120.  Infinite  geometrical  series.  If  the  number  of  terms 
of  a  geometrical  series  is  unlimited,  it  is  called  an  infinite 
geometrical  series. 


PEOGEESSIONS  221 

In  the  progression  2,  4,  8,  •  •  •  the  ratio  is  positive  and 
greater  than  1,  and  each  term  is  greater  than  the  term 
preceding  it.  Such  a  progression  is  said  to  be  increasing. 
Obviously  the  sum  of  an  unlimited  number  of  terms  in 
such  an  increasing  geometrical  progression  is  unlimited. 
If  r  >1  (read  "r  is  greater  than  1"),  the  sum  can  be  made 
as  large  as  we  please  by  taking  enough  terms. 

In  the  progression  2,  1,  i,  i,  i,  Jg,  •  •  .  the  ratio  is 
positive  and  less  than  1,  and  each  term  is  less  than  the  term 
preceding  it.  Such  a  progression  is  said  to  be  decreasing. 
Though  the  number  of  terms  of  such  a  geometrical  pro- 
gression be  unlimited,  the  sum  of  as  many  terms  as  we 
choose  to  take  is  always  less  than  some  definite  number. 

The  sum  of  the  first  three  terms  of  the  series  2  +  1+^+:^+^ 
+  •  •  •  is  3^ ;  of  four  terms,  is  3| ;  of  five  terms,  is  3| ;  of  six  terms, 
is  3^f ;  of  seven  terms,  is  3|^,  etc.  Here  for  any  number  of  terms 
the  sum  is  always  less  than  4. 

The  limit  of  the  sum  of  the  series  2  +  l4-i+i+i+t's"+*" 
can  be  seen  by  reference  to  the  following  diagram  : 


D E^FGX 


i— -r-i 


Here  the  terms  of  the  above  series  are  represented  by  intervals  on  the 
line  AXj  which  is  four  units  in  length;  that  is,  AB  =  2,  BC —  1, 
CD  =  I,  DE=  1,  EF=  I,  FG  =  ^V  etc. 

A  study  of  the  several  intervals  reveals  the  fact  that  as  we  pass 
from  A  toward  X  each  new  interval  is  just  one  half  of  the  then  unused 
portion  of  AX.  It  follows,  then,  that  the  distance  laid  off  gradually 
approaches,  though  it  never  equals,  AX,  for  at  no  time  is  an  interval 
adjoined  which  is  more  than  one  half  of  that  which  then  remained. 
Therefore  there  will  always  be  an  unused  interval  between  the 
extremity  of  the  last  interval  used  and  the  point  X. 

Since  the  sum  of  the  intervals  laid  off  approaches  but  never  quite 
equals  the  interval  AX,  we  have  in  the  above  diagram  an  illustration 
of  the  fact  that  the  sum  of  the  series  2+l+i  +  ;|^  +  ^  +  TV+  ••• 
approaches,  yet  always  remains  less  than,  4. 


222  SECOND  COURSE  IN  ALGEBRA 

a  —  ar^ 


The  formula  for  the  sum  of  a  geometric  series  S^  — 
may  be  written  as  the  difference  of  two  fractions, 

1  —  r      1  —  r 

For  the  series  2  + 1  +  J  +  |  +  •  •  •  it  follows  that 

Now  (1)2=1,  (1)8=1,  (-1-)".=  ^^,  .(J/  =  3:V  Conse- 
quently  (|)^  becomes  very  small  if  n  is  taken  very  great. 
Therefore  2(|)^,  the  numerator  of  the  last  fraction  in  (3), 
decreases  and  approaches  zero  as  n  increases  without  limit. 
Since  the  denominator  of  the  fraction  remains  ^  while 
the  numerator  approaches  zero,  the  value  of  the  fraction 
decreases  and  approaches  zero  as  n  increases.  Then  if 
>S^oo  denotes  S^  where  n  has  increased  without  limit,  we 
may  write  9 

Soo    approaches  »  or  4. 

2 

In  general,  if  r  <  !(''  if  r  is  numerically  less  than  1"),  the 

numerical  value   of  fraction approaches  zero   as  n 

1  — r 

increases  without  limit.   Under  such  conditions  the  formula 

S,,  = -z ■    becomes    aSoo  =  -z • 

1— rl— r  \  —  r 

This   means   that   for   r  numerically    less   than   1,    S^ 

approaches •>   but  for  any  definite  value  of  n  it  is 

always  numerically  less  than  this  number. 

Hence  whenever  we  speak  of  the  sum  of  such  a  series 
we  mean  the  limit  which  the  sum  approaches  as  n  increases 
indefinitely. 


PROGRESSIONS 


223 


EXERCISES 

Find  the  sum  of  the  following  series : 

8.  5  +  V54-I4- 


1.  4  +  1  +  1  +  .  . 
Solution.  So 

Substituting,    So 


i-i 


9-  l  +  ;  +  3  + 


(x>l) 


2.  l  +  ^  +  i  +  .... 

3.  3  +  (-l)  +  i  +  ---. 

4.  5-|-(-2)+A  +  .... 

5.  2  + V2  +  H . 

_  5a      5a 

6.  5a  +  — +  —  +.... 

7.  l  +  x  +  a;2+  ....   (x<l) 


10.  .121212-. 

Hint.    .121212  = 

11.  .666... 

12.  .151515.... 

13.  .3232.... 

14.  25.2727... 

15.  .71515.... 

16.  .3108108.... 


17.  A  flywheel  whose  circumference  is  5  feet  makes  80 
revolutions  per  second.  If  it  makes  99%  as  many  revolu- 
tions each  second  thereafter  as  it  did  the  preceding  second, 
how  far  will  a  point  on  its  rim  have  moved  by  the  time  it  is 
about  to  stop  ? 

Note.  In  the  study  of  geometrical  progressions  we  have  seen 
that  the  sum  of  the  infinite  series  1  +  a;  +  a:^  +  a;^  +  . . .  is  a  definite 
number  when  x  has  any  value  less  than  1.  But  it  has  no  finite  value 
when  X  is  equal  to  or  greater  than  1 ;  that  is,  we  have  an  expression 
which  we  cannot  use  arithmetically  unless  x  has  a  properly  chosen 
value.  If  we  were  studying  some  problem  which  involved  such  a 
series,  it  would  be  a  matter  of  the  most  vital  importance  to  know 
whether  the  values  of  x  under  discussion  were  such  as  to  make  the 
series  meaningless. 

This  question  of  distinguishing  between  expressions  which  con- 
verge (that  is,  the  sum  of  whose  terms  approaches  a  limit)  and  those 
which  do  not  has  an  interesting  history.  Newton  and  his  followers  in 
the  seventeenth  century  dealt  with  infinite  series  and  always  assumed 


224  SECOND  COURSE  IN  ALGEBRA 

that  they  converged,  as,  in  fact,  most  of  them  did.  But  as  more 
complicated  series  came  into  use  it  became  more  difficult  to  tell  from 
inspection  whether  they  meant  anything  or  not  for  a  given  value 
of  the  variable. 

It  was  not  until  the  beginning  of  the  nineteenth  century  that 
Gauss,  Abel,  and  Cauchy,  in  Germany,  Norway,  and  France  respec- 
tively, began  to  study  this  subject  effectively  and  to  devise  far- 
reaching  tests  to  determine  the  values  of  x  for  which  certain  series 
converge  to  a  finite  limit.  It  is  said  that  on  hearing  a  discussion 
by  Cauchy  in  regard  to  series  which  do  not  always  converge,  the 
astronomer  La  Place  became  greatly  alarmed  lest  he  had  made  use 
of  some  such  series  in  his  great  work,  **  Celestial  Mechanics."  He 
hurried  home  and  denied  himself  to  all  distractions  until  he  had 
examined  every  series  in  his  book.  To  his  intense  satisfaction  they 
all  converged.  In  fact  it  has  often  been  observed  that  a  genius  can 
safely  take  chances  in  the  use  of  delicate  processes,  which  seem 
very  foolish  and  unsafe  to  a  man  of  ordinary  ability. 

MISCELLANEOUS  EXERCJSES 

1.  The  digits  of  a  certain  three-digit  number  are  in  geo- 
metrical progression.  The  sum  of  the  digits  is  13.  The  number 
divided  by  the  sum  of  its  digits  gives  a  quotient  of  10  and  a 
remainder  of  9.    Find  the  number. 

2.  What  per  cent  of  the  eleventh  term  of  the  progression 
1,  2,  4,  8,  .  .  •  is  the  eleventh  term  of  1,  101,  201,  •  •  •  ? 

3.  Compare  the  sum  of  the  first  ten  terms  of  tfce  first  pro- 
gression in  Exercise  2  with  the  sum  of  the  first  ten  terms  of 
the  second  progression. 

4.  What'  meaning  may  be  attached  to  (a)  $600.(1.06)'? 
ih)  $500  .  (1.05)^?  (c)  $500  .  (1.03)«? 

5.  What  will  $100  amount  to  in  three  years,  with  interest 
at  4%,  compounded  annually?  in  five  years  ? 

6.  What  will  $100  amount  to  in  two  years,  with  interest  at 
6%,  compounded  semiannually  ?  in  three  years  ? 


PROGRESSIONS  225 

7.  For  eacli  of  ten  successive  years  a  man  saves  $200  from 
his  salary  and  places  it  in  a  savings  bank  where  it  draws  4<^ 
interest,  compounded  annually.  Find  the  total  amount  of  his 
savings  at  the  time  of  his  fifth  annual  deposit.  Indicate  (with- 
out multiplication)  the  amount  of  his  deposit  at  the  beginning 
of  the  eleventh  year. 

8.  A  loan  of  S  dollars  is  to  be  repaid  in  four  equal  annual 
payments  ot  p  dollars  each.    Find  j9  if  money  is  worth  r%. 

Solution.    The  sum  due  at  beginning  of  second  year  is 

Sa  +  r)-p,  (1) 

The  sum  due  at  beginning  of  third  year  is 

[^(l  +  r)-p](l+r)-p.  (2) 

The  sum  due  at  beginning  of  fourth  year  is 

{[S(l  +  r)  -  ;>]  (1  +  r)  -  ;.  }  (1  +  r)  -  p.  (3) 

The  sum  due  at  beginning  of  fifth  year  is 

[{['^(1  +  0  -i^]  (1  +  ^)  -i>}  a  +  0  -i>]  a  +  r)  -p.     (4) 

By  the  conditions  of  the  problem,  (4)  =  0,  for  all  the  debt  has 
then  been  paid.    Setting  (4)  equal  to  zero  and  simpUfying, 

S(l  +  ry-p{l  +  ry-p(l-{-ry-p{l+r)-p  =  0.         (5) 

Solving  (5)  ior  p, 

S(l  +  ry 

(1  +  ry  +  (1  +  ry  +  (1  +  r)  +  1 
r 

But  the  denominator  in  (6)  is  a  geometrical  series  whose  sum 

by  the  formula  *S„,  =  is  -'^ ^ ■- 

1—  r  .        r 

Substituting  this  last  value  for  the  denominator  of  (6), 

^       (1  +  ry-l  ^^ 

In  the  general  case,  if  we  have  n  annual  payments,  the  exponent 

Sr(l  +  r^'^ 
4  in  (7)  would  be  replaced  by  n,  and  then  p  =  -——^ ^ — 


p=..  ,„.s,.rv:;^ ,-  w 


226 


SECOND  COUESE  IN  ALGEBRA 


9.  A  loan  of  flOOO  is  to  be  repaid  in  three  equal  annual 
payments,  interest  at  6%.    Find  the  payment  required. 

10.  A  loan  of  |2000  bearing  interest  at  5%  is  to  be  repaid 
in  five  equal  annual  payments.    Find  the  payment  required. 

11.  The  machinery  of  a  certain  mill  cost  |10,000.  The  owner 
figures  that  the  machinery  depreciates  10%  in  value  each  succes- 
sive year.  What  was  the  estimate  on  the  value  of  the  machinery 
at  the  end  of  the  sixth  year  ? 

12.  A  vessel  containing  wine  was  emptied  of  one  third  of 
its  contents  and  then  filled  with  water.  This  was  done  four 
times.  What  portion  of  the  original  contents  was  then  in  the 
vessel  ?  (J 

13.  In  the  adjacent  figure  the  tri- 
angle DEF  is  formed  by  joining  the 
mid-points  of  AB,  BC,  and  CA  respec- 
tively. Triangles  GHI  and  KLAI  are 
formed  in  like  manner.  If  ABC  is 
equilateral,  prove  that  the  successive 
perimeters  of  the  triangles  form  a 
geometrical  progression,  li  AB  =  5,  find  the  sum  of  the 
perimeters   of   all   the   triangles   which   may    be   so   formed. 

14.  Each  square  in  the  adjacent  fig- 
ure except  the  first  is  formed  by  join- 
ing the  mid-points  of  the  ftquare  next 
larger.  It  AB  =  4:,  show  that  the  perim- 
eters of  the  squares  form  a  decreasing 
geometrical  progression.  Find  the  sum 
of  the  perimeters  of  all  the  squares 
which  may  be  so  drawn? 


CHAPTER  XVII 
THE  BINOMIAL  THEOREM 

121.  Powers  of  binomials.  The  following  identities  are 
easily  obtained  by  actual  multiplication : 

(^  j^hy  =  d^-\-2ah  +  h\  (1) 

(a  Jrhy  =  a^  +  Z  a%  +  Zal^  +  lA  (2) 

(a  4-  6)4  =  ^4  +  4  a%  +  6  a.262  +  4  aS^  _^  h\  (3) 

(a  +  6)5  =  ^5  _^  5  ^46  +10  ^352  + 10  ^263  _^  5  a64  +  65.  (4) 

If  a  -f-  6  is  replaced  by  a  —  h,  the  even-numbered  terms 
in  each  of  the  preceding  expressions  will  then  be  negative 
and  the  odd-numbered  terms  will  be  positive. 

122.  The  expansion  of  (a+by.  The  form  of  the  expan- 
sion for  the  general  case  will  now  be  mdicated : 

The  first  term  is  a"  and  the  last  is  6". 
The  second  term  is  -h  na"~^b. 

The  exponents  of  a  decrease  hy  1  in  each  term  after  the 
first. 

The  exponents  of  b  increase  hy  1  in  each  term  after  the 
second. 

The  product  of  the  coefficient  in  any  term  and  the  'exponent 
of  a  in  that  term.,  divided  hy  the  exponent  of  b  increased  hy  i, 
gives  the  coefficient  of  the  next  term. 

The  sign  of  each  term-  is  -\-  if  a  and  b  are  positive ;  the 
sign  of  each  even-numhered  term  is  —  if  b  alone  is  negative, 

227 


228  SECOND  COURSE  IN  ALGEBRA 

According  to  the  above  statement  we  have 

(a  +  b)"  =  a"  +  7 a"-^b+  "^f  ~  ^  a"-^bi' 
1     •  1*2 

n(n-l)(n-2) 

■^        1.2.3       ''""^  +  .  .  .  +  &^. 

This  expresses  in  symbols  the  law  known  as  the  binomial 
theorem.  The  theorem  holds  for  all  positive  values  of  n  and 
also,  with  certain  limitations,  for  negative  values. 

Note.  The  coefficients  of  the  various  terms  in  the  binomial 
expansion  are  displayed  in  a  most  elegant  form  as  follows : 

1 

1  1 

12  1 

13  3  1 

14  6  4  1 


In  this  arrangement  each  row  may  be  derived  from  the  one  above  it 
by  observing  that  each  number  is  equal  to  the  sum  of  the  two  num- 
bers, one  to  the  right  and  the  other  to  the  left  of  it,  in  the  line  above. 
Thus  4=1  +  3,  6  =  3  +  3,  etc.  The  next  line  is  1  5  10  10  5  1. 
The  successive  lines  of  this  table  give  the  coefficients  for  the  expan- 
sions of  (a  +  &)"■  for  the  various  values  of  n.  The  numbers  in  the  last 
line  of  the  triangle  are  seen  to  be  the  coefficients  when  n  =  4 ;  the 
next  line  would  give  those  for  n  =  5.  This  array  is  known  as  Pascal's 
triangle,  and  was  published  in  1665.  Tt  was  probably  known  to 
Tartaglia  nearly  a  hundred  years  before  its  discovery  by  Pascal. 

ORAL  EXERCISES 
What  is  the  second  term  in  the  expansions  of  Exercises  1-4  ? 
1.  (a  +  by\         2.  (a-b)"".         3.  (a-\-by\         4.  (a  -  bfK 

Assuming  that  the  terms  in  Exercises  5-10  occm-  in  an  ex- 
pansion of  the  binomial  a  -\-  b,  find  (a)  the  exponents,  (b)  the 
coefficient  in  the  next  following  term  of  the  expansion. 

5.  10a%\  7.  15a*b\  9.  252  a^b^. 

6.  Sab\  8.  11  .  5  .  9a'b\  10.  20  a^b\ 


THE  BINOMIAL  THEOKEM  229 

EXERCISES 
Expand  by  the  rule : 

1.  (a  +  b)\  3.  (a  +  1)'.  5.  {a  +  3)«. 

2.  {a  -  If.  4.  (a  +  2)^  6.  (2  -  af. 

Obtain  the  first  four  terms  of  the  following  : 

7.  (a  +  by\       8.   (a  +  b)"".        9.  (a  +  1)^.        10.  (a  -  2)2^. 

Expand : 

11.  (a^  +  2by. 

Hints.    To  avoid  confusion  of  exponents  first  write 
(a2)&  +      (a2)4  (2  6)1  +      (^2)3  (2  6)2  +      (a2)2  (2  6)3  +       (a2)i  (2  6)*  +  (2  6)5. 

Then  in  the  spaces  left  for  them  put  in  the  coefficients  according  to 
the  rule  of  this  section. 

Finally,  expand  and  simplify  each  term. 

12.  («.^  +  2)«.     13.  (a^-24)'.     14- (^ J-     15.  (^)'. 

Obtain  in  simplest  form  the  first  four  terms  of  the  following  ; 

16.  (a^  +  3  b)"".  /a^      2^Y 

17.  (a'^-Sby.  '  ^^'        ^*/ 


19.  (t  +  -^)  •  23.  [^  +  ^J. 

20.  (^-^)-  24.(1  +  -J 


■■&-sf'      -('-;: 


25.  Write  the  first  six  terms  of  the  expansion  ol  (a  +  by\ 
and  evaluate  it  for  n  =  1,  n  =  2,  n  =  S,  n  =  4:.  How  does  the 
number  of  terms  compare  with  n  ?  What  is  the  value  of  each 
coefficient  after  the  (t^  -h  l)th  ?  Why  does  not  the  expansion 
extend  to  more  than  five  terms  when  n  =  4^? 


230  SECOND  COURSE  IN  ALGEBRA 

Compute  the  following  to  two  decimal  places.  (In  each  case 
carry  the  computation  far  enough  to  be  certain  that  the  terms 
neglected  do  not  affect  the  second  decimal  place.) 

26.  (l.l)^*'.  29.  (.98)". 

Hint.   (1.1)1°  z=  (1  +  .l)io  etc.  Hint.   (.98)"  =  (1  -  .02)"  etc. 

27.  (5.2)«.  30.  (4.9)^ 

28.  (1.06)«.  31.  (2.9)1 

123.  Extraction  of  roots  by  use  of  the  binomial  expansion. 
The  expansion  in  section  122  may  be  verified  for  any  par- 
ticular integral  value  of  n  without  difficulty  by  direct  mul- 
tiplication, as  in  section  121.  But  if  n  has  a  negative  or 
fractional  value,  a  laborious  proof  is  required  to  show  that 
the  expansion  is  still  valid  when  a  is  numerically  greater 
than  h.  Since  none  of  the  factors  of  the  coefficients,  as  n, 
n  —  1,  n  —  2,  vanish  for  fractional  or  negative  values  of  n^ 
it  appears  that  for  such  exponents  the  expansion  is  an 
infinite  series. 

For  example, 

(a  +  h)^  =  a^  +  \arh  -\'a~h'^  +  -^^  .  a"  V+  •  •  .. 

By  giving  n  the  values  |-  or  J,  one  can  compute  the 
square  root  or  the  cube  root  of  a  number  to  any  required 
degree  of  accuracy. 

In  such  computations  it  is  desirable  to  let  the  number 
which  corresponds  to  a  in  the  binomial  exceed  the  one 
corresponding  to  h  by  as  much  as  possible  and  at  the  same 

time  to  have  a**  an  integer. 

Note.  'The  process  of  extracting  the  square  root  and  even  the 
cube  root  by  means  of  the  binomial  expansion  was  familiar  to  the 
Hindus  more  than  a  thousand  years  ago.  The  German  Stifel  (1486- 
1567)  stated  the  binomial  theorem  for  all  powers  up  to  the  seven- 
teenth, and  also  extracted  roots  of  numbers  by  this  method. 


THE  BINOMIAL  THEOEEM  231 

EXERCISES 

Eind  the  first  four  terms  of  the  following : 

1.  (l  +  x)\  3.  (S-x)i.  5.  (2  +  x)^. 

2.  (2-\-x)K  4.  (l  +  x)k  6.  (S-x)^. 

Eind  to  at  least  three  decimals  by  the  binomial  theorem : 
7.  (27)i 

Solution.    (27)^  =  (25  +  2)^ 

=  25^+  i.25-2.  2-^-25-t.22 

+  V^.25-^.23 

=  5  +  .2  -  .004  +  .00016  .  •  •  =  5.196 +  . 

It  is  proved  in  more  advanced  books  that  when  the  terms  of 
an  infinite  series  are  alternately  plus  and  minus,  and  each  term  is 
numerically  less  than  the  preceding  one,  the  value  of  the  entire  sum 
from  a  given  term  on  cannot  exceed  that  term.  This  fact  renders 
these  so-called  "  alternating  series "  especially  convenient  for  com- 
putation, since  a  definite  limit  of  error  is  known  at  each  stage  of 
the  computation.    In  this  example  the  error  cannot  exceed  .00016. 

8.  (lT)k       9.  (28)1       10.  (38)i        11.  (78)^       12.  (125)i 

13.  (61)i 

Solution.    (61)^  =  (64  -  3)i 

=  64i  -  J-  •  64- 1 .  3  -  ^  .  64-  ^  .  3^ 

-/r-64-t.33-f,-. 
=  ^  ~  T^?r  —  tttV?  —  t-^At^w  +  . . . 
=  4  -  .0625  -  .00097  -  .00000259  •  •  •  =  3.93653  -  . 

Here  three  terms  give  the  result  to  five  figures. 

14.  (79)i 

Hint.    (79)^  =:  (81  -  2)i  =  81^-  -|  •  81"^  •  2+  .  • .. 

Here  (81  —  2)^  yields  more  accurate  results  with  fewer  terms  than 
does  (64  +  15)i. 

15.  (28)i  16.  (66)i  17.  (30)i  18.  (700)i 


232  SECOND  COUESE  IN  ALGEBRA 

124.  The  factorial  notation.  The  notation  5 !  or  [5  sig- 
nifies 1  .  2  .  3  .  4  .  5,  or  120.    Also  4!  =1  .  2  .  3  •  4  =  24. 

In  general,  t^  !  =  1  •  2  •  3  •  4  . .  •  (w  —  2)  (/i  —  1)  n. 
The  symbol  nl  or  [w  is  read  "factorial  ti." 

In  the  factorial  notation  the  denominators  of  the  fourth  and 
fifth  terms  of  the  expansion  of  (a  +  by  become  3  1  and  4  I  respec- 
tively (see  formula,  p.  228). 

EXERCISES 

Evaluate : 

1.  6!.  3.  5!. 2!.  5.  4!  -3!  •  2!. 

2.  4!. 3!.  4.  6!--2!.  6.  ti!  ^(/i -1)!- 

^     ,     ^    n(n-l)(n-2) -.-  (n-r-^2)       ^ 

Evaluate  -^ ^ — ,    \^, — ^'  when: 

(r-l)l 

7.  n=7,r  =  5.  10.  n  =  20,  r  =15. 

8.  71=15,  r  =  S.  11.  n=lS,r=17. 

9.  71  =  21,  r  =  12.  12.  n  =  10,  r  =  11. 

125.  The  rth  term  of  (a  +  &)".  According  to  the  bmomial 
theorem  the  fifth  term  of  the  expansion  on  page  228  is 

n(n-l)(n-  2) (n -  S)a''-^^ 

.    "^  4! 

• 

If  we  note  carefully  this  term  and  the  directions  on 
page  227,  we  can  write  down,  from  the  considerations 
that  follow,  any  required  term  without  writing  other 
terms  of  the  expansion. 

The  denominator  of  the  coefficient  of  the  fifth  term  is  4  I 
From  the  law  of  formation  the  denominator  of  the  sixth 
term  would  be  5 !,  of  the  seventh  term  6 1,  etc.  Conse- 
quently in  the  rth  term  the  denominator  of  the  coefficient 
would  be  (r  —1)1. 


THE  BINOMIAL  THEOEEM  233 

The  numerator  of  the  coefficient  of  the  fifth  term  contains 
the  product  of  the  four  factors  ?^(7^— l)(/i— 2)(?i— 3). 
The  numerator  of  the  sixth  term  would  contain  these  four 
and  the  additional  factor  7i  —  4.  Similarly,  the  last  factor 
in  the  numerator  of  the  seventh  term  would  be  n  —  5,  etc. 
Hence  the  last  factor  in  the  rth  term  would  be  7i  —  (r  —  2), 
and  the  numerator  of  the  coefficient  of  the  rth  term  is 
7i(n-l)(^-2)(^-3)  ...  (w-r  +  2). 

The  exponent  of  a  in  the  fifth  term  is  n  —  4,  and  in  the 
sixth  term  it  would  be  n  —  5,  etc.  Therefore  in  the  rth 
term  the  exponent  of  a  is  n  —  (r  —  1),  or  n  —  r  + 1. 

The  exponent  of  b  in  the  fifth  term  is  4,  in  the  sixth 
term  is  5,  etc.  Therefore  in  the  rth  term  the  exponent  of 
h  is  r  —  1. 

The  sign  of  any  term  of  the  expansion  (if  n  is  a  posi- 
tive integer)  is  plus  if  the  binomial  is  a  +  b.  If  the 
binomial  is  a  —  b,  the  terms  containmg  the  odd  powers  of 
b  will  be  negative.  In  other  words,  the  sign  in  such  cases 
depends  upon  whether  the  exponent  r  —  1  is  odd  or  even. 

Hence  the  rth  term  (r  not  equal  to  1)  of  («  +  b^^^  equals 

n(n-l)(/2-2)(n-3).  •  .(n-r+2) 


(r-1)! 


.a"-'+^lf-K       (1) 


If  we  wanted  the  twelfth  term,  we  would  in  using  (1) 
substitute  12  for  r. 

EXERCISES 

Write  the  indicated  terms  : 

1.  Fifth  term  of  (a  +  bf^.  2.   Sixth  term  of  (a  +  bf. 

Solution.    Substituting  10  for  n 


,-.       •    .^    f         1    /ix    •  3.  Fourth  term  of  (a  +  ^)^. 
and  o  for  r  m  the  formula  (1)  gives  ^      '      ^ 

lQ-9-8-7        ^  10-9 -8 -7  4.   Seventh  term  of  (a  -  bf 
4!                       4-3.2 

=  210  a«&*.  5.  Eighth  term  of  (a  —  bf^ 


234  SECOKB  COURSE  IN  ALGEBRA 

6.  Fourth  term  of  (a +  -)   •      8.  Sixth  term  of  ( ]    • 

7.  Fifth  term  of  (a^  -  hf".        9.  Middle  term  of  {x"  -  xf. 

10.  Seventh  term  of  (-; )    • 


(v-.-^r 


11.  Fifth  term  of 

Find  the  coefficient  of : 

12.  x^  in  (1  +  xy.  14.  x^^  in  (x^  -f- 1)'^ 

13.  x^  in  (1  +  xY'  15-  x^'  in  (^^  -  »"')" 

Note.  The  binomial  theorem  occupies  a  remarkable  place  in  the 
history  of  mathematics.  By  means  of  it  Napier  was  led  to  the  dis- 
covery of  logarithms,  amd  its  use  was  of  the  greatest  assistance  to 
Newton  in  making  his  most  wonderful  mathematical  discoveries. 
But  to-day  the  results  of  Newton  and  of  Napier  are  explained  with- 
out even  so  much  as  a  mention  of  the  binomial  theorem,  for  simpler 
methods  of  obtaining  these  results  have  been  discovered. 

It  was  Newton  who  first  recognized  the  truth  of  the  theorem,  not 
only  for  the  case  where  n  is  a  positive  integer,  which  had  long  been 
familiar,  but  for  fractional  and  negative  values  as  well.  He  did  not 
give  a  demonstration  of  the  general  validity  of  the  binomial  develop- 
ment, and  none  even  passably  satisfactory  was  given  until  that  of 
Euler  (1707-1783).  The  first  entirely  satisfactory  proof  of  this 
difl&cult  theorem  was  given  by  the  brilliant  young  Norwegian  Abel 
(1802-1829). 


CHAPTER  XVIII 

RATIO,  PROPORTION,  AND  VARIATION 

126.  Ratio.  The  ratio  of  one  number  a  to  o,  second 
number  b  is  the  quotient  obtained  by  dividing  the  first 

by  the  second,  or  -.    The  ratio  of  ^  to  5  is  also  written 

J.  ^ 

a:  0. 

It  follows  from  the  above  that  all  ratios  of  two  numbers 

are  fractions  and  all  fractions  may  be  regarded  as  ratios. 

Thus  -'  — '  — '  and  — — :  are  ratios  as  well  as  fractions. 

5    ^d   c-d  V3 

Since  ratios  like  the  above  are  fractions,  operations 
which  may  be  performed  on  fractions  may  be  performed  on 
these  ratios.  Hence  the  value  of  a  ratio  is  not  changed 
by  multiplying  or  dividing  both  numerator  (antecedent) 
and  denominator  (consequent)  by  the  same  number. 

a 

Thus  -  = and     -  =  t- 

h       b'  X  ^2. 

y 

EXERCISES 

Simplify  the  following  ratios  by  considering  them  as  frac- 
tions and  reducing  the  fractions  to  lowest  terms : 

,2 


-i^-^^^-i^*^^^ 


2.  1  kilometer  :  1  mile.    (1  kilometer  =  .62  miles.) 

3.  1  liter  :  1  quart.     (1  liter  =  .001  cubic  meter ;  1  quart  = 
^^  cubic  inches  ;  and  1  meter  =  39.4  inches.) 

235 


236  SECOND  COURSE  IN  ALGEBRA 

4.  A  city  lot  100  x  160  feet :  1  acre.  (1  acre  =  43,560  square 
feet.) 

5.  Area  of  printed  portion  of  this  page  :  total  area  of 
the  page. 

6.  If  |24,000  is  divided  between  two  men  so  that  the 
portions  received  are  to  each  other  as  5:7,  how  much  does 
each  receive? 

Hint.    Let  5  x  and  7x  be  the  required  parts. 

7.  Separate  690  into  four  parts  which  are  to  each  other 
as  2  :  5  :  7  :  9. 

X  X  -\-  S 

8.  Show  that -< ;:  if  x  is  positive. 

X  -\-  2     X  -^  5 

Hint.  Reduce  the  given  fractions  to  respectively  equivalent  fractions 
having  a  common  denominator,  then  compare  the  numerators  of  the 
fractions  so  obtained. 

9.  Arrange  the  ratios  3 : 4  and  7 : 9  in  decreasing  order  of 
magnitude. 

127.  Proportion.  A  proportion  is  a  statement  of  equality 
between  two  ratios.  Four  numbers,  a,  h,  <?,  and  d,  are  in 
proportion  if  the  ratio  of  the  first  pair  equals  the  ratio  of 
the  second  pair. 

This  proportion  is  written 

a:  b  =  c:  a     or     r  =  -: • 
b     a 

Though  both  forms  are  equations,  the  second  is  the 
more  familiar  one  and  for  this  reason  is  preferable. 

'  Note.  By  the  earlier  mathematicians  ratios  were  not  treated  as 
if  they  were  numbers,  and  the  equality  of  two  ratios  which  we 
know  as  a  proportion  was  not  denoted  by  the  same  symbol  as  other 
kinds  of  equality.    The  usual  sign  of  equality  for  ratios  was  : : ,  a 


EATIO,  PROPOETION,  AND  VARIATION      237 

notation  which  was  introduced  by  the  Englishman  Oughbred  in 
1631  and  was  brought  into  common  use  by  John  Wallis  about  1686. 
The  sign  =  was  used  in  this  connection  by  Leibnitz  (1646-1716) 
in  Germany,  and  by  the  Continental  writers  generally,  while  the 
English  clung  to  Oughtred's  notation. 

ci       c 
In  the  proportion  t  =  -^  the  first  and  fourth  terms  (a,  cZ) 
0       ci 

are  called  the  extremes,  and  the  second  and  third  terms 

(^,  (?)  are  called  the  means. 

128.  Mean  proportional.   A  mean  proportional  between  two 

numbers  a  and  h  is  the  number  m  if  —  =  —  •  It  follows 
that  Tn^  =  ab^  or  m  =  ±  Va6. 

129.  Third  proportional.  A  third  proportional  to  two  num- 
bers a  and  h  is  the  number  t  \i  -  =  -- 

0       t 

130.  Fourth  proportional.    A  fourth  proportional  to  three 

ci       c 
numbers  a,  5,  and  c  is  the  number  /  if  -  =  — . 

131.  Test  of  a  proportion.  Since  a  proportion  is  an 
equality  between  two  ratios  (fractions),  it  is  therefore 
an  equation.  Hence  any  operation  which  may  he  performed 
on  an  equation  may  he  performed  on  a  proportion,  (See 
Axioms,  p.  16.) 

Thus,  in  the  proportion  —  =  -  both  members  may  be  multiplied 
h      d 

by  hd,  giving  ad  =  he.  Here  the  first  member  is  the  product  of  the 
extremes  of  the  proportion,  and  the  second  member  is  the  product 
of  the  means. 

Therefore  in  any  proportion  the  product  of  the  extremes  equals  the 
product  of  the  means. 

132.  Proportions  from  equal  products.  The  numbers 
which  occur  in  a  pair  of  equal  products  may  be  used  in 
various  ways  as  the  terms  of  a  proportion. 


238  SECOND  COURSE  IN  ALGEBRA 

Thus,  if  ad  =  be, 

we  may  write  either 


a      c  ^  _  ^ 


Proof,   li  a-  d  =  h-  c  is  divided  by  hd,  we  obtain 


ad       he  a       e  /^\ 

—  =  — ,     or     -  =  —  (1) 

hd      hd  h      d 


li  a  '  d  =  b  '  c  is  divided  by  ed,  we  obtain 

c      d' 


(2) 


If  the  means  in  (1)  are  interchanged,  (2)  is  obtained. 
This  process  of  obtaining  (2)  from  (1)  is  called  alternation. 

li  a-  d  =  b-  c  is  divided  by  ac,  we  obtain 

-  =  -■  (3) 

a      e 

If  the  fractions  in   (1)  are  inverted,  (3)  is  obtained. 
This  process  of  obtaining  (3)  from  (1)  is  called  inversion. 

EXERCISES 

1.  Find   a   mean   proportional   between:    (a)  3  and   27; 

(b)  y  and  -;   (c)  —  and  -;   (d)  ^  and  xy. 

2.  Find  a  third  proportional  to:    (a)  9  and  6;    (b)  180 
and  60;   (c)  216  and  36. 

3.  Find  the   fourth  proportional  to :    (a)  14,  10,  and  7 ; 

(b)  27,  3,  and  36;  (c)  96,  12,  and  8. 

4.  Form   three   proportions   from  each  of   the  following 
equations  :    (a)    5  x  =  ^  y  ;    (b)    (a  4-  3)  •  2  =  (a  +  1)  •  3  ; 

(c)  dip'  —  y^  —  T^  —  s\ 

2      10  2      20 

5.  Write  by  alternation:  (a)  -  =  —  5  (b)  -  =  — 

6.  Write  by  m version:  («)-  =  -—?  (o)  -  =  t^tt- 

•^  ^^6      12^^ic30 


RATIO,  PROPOETION,  AND  VAEIATION      239 

a       c 
If  -  =  -,  prove  the  following  and  state  the  corresponding 


theorems  in  words 


7. 

a 

h 

G 

d 

h 

d 

8. 

a 

c 

9. 


10. 


11. 


b''      d'' 
-\a  _  Vc 

a  -\-  b       c  -\-  d 


b  d 

Hint.   Add  1  to  each  member 


.a      c 

of  -  =  — 

b      d 


12. 


a  -i-  b       c  -\- 


a 


by  inversion 


Hint.    Write  - 

b      d 
and  apply  hint  of  Exercise  11. 


13. 
14. 
15. 


a  —  b  _c  —  d 
b      ^      d 

a  —  b       G  —  d 


a 

a-\-b 


G 

c-\-d 


a  —  b       c  —  d 


The  proportions  given  in  Exercises  11,  13,  and  15  are  said 
to  be  derived  from  a  :b  =  g:  d  by  addition,  subtraction,  and 
addition  and  subtraction  respectively. 


If  -[=-,'>  show  that  the  following  equalities  are  true : 
0       di 


^^    5a      5c 

,«    2a      2g 
7b       Id 


18. 


19. 


P 


ac 
bd 


5a  -\-b  _5g  -\-  d 
5a  —  b      5g  —  d 


20. 
21. 
22. 
23. 


b^ 


d^ 


b' 
a''-7P 


d^ 
c'-ld^ 

2 -(^2 


2ab  2Gd 

7a2_3^2      7c2_3^2 


5ab 


5Gd 


24. 


a'-^-ab-Jrb^  _a'-  ah  -f-  b'' 
g'  +  cd  -\-d''~  c'-Gd-\-d'' 


240  SECOND  COUKSE  IN  ALGEBRA 

25.  In  the  proof  which  follows  give  the  reason  for  each 
step  and  state  the  result  as  a  theorem : 

o_c_e  a-\-c-\-e  _a  _c  _e 

l^d^f        ^^'     b  +  d+f~b~d~f 
Proof.    Setting  each  of  the  given  ratios  above  equal  to  r, 

-  =  r,     -  =  r,     and     -  =  r.  (1) 

Then  from  (1),  a  =  hr,     c  =  d?^     e  =fr.  (2) 

Adding  in  (2),  a  +  c  +  e  =  br -]- dr +fr.  (3) 

Factoring  in  (3),  a  +  c  +  e  =  (h  +  d +fy.  (4) 

Therefore  1^7^  =  ''  ^^^ 


Hence  by  (1)  and  (5), 


a-\-c  +  e_a_c_e 

bTJTf  ~b~d~f 


26.  If  T  =  3  =  -  5  show  that  =  -    or  -• 

Solve,  using  theorems  of  proportion  : 

27.  2a^:(ic  +  8)  =  10:3.         29.  5  :  4  =  (.t  -  3):(£r  -  4). 

28.  25:x  =  x:169.  30.  (15  +  x):(15  -  «)  =  13  :17. 

31.  (10  +  £c):(20  +  3cc)  =  (10-ic):-3ir. 

32.  V^  +  2^^+1 

■y/x-2      x-7 
33.  Show  that  the  mean  proportional  between  two  numbers 
is  the  geometric  mean  between  these  numbers. 

PROBLEMS 

1.  The  surface  of  a  sphere  is  4  7r72^    If  S  represents  the 
surface  of  a  sphere,  R  its  radius,  and  D  its  diameter,  show 

2.  Find  the  ratio  of  the  surfaces  of  two  spheres  whose 
radii  are  in  the  ratio  1:10, 


KATIO,  PEOPORTION,  AND  VAKIATION      241 

3.  If  the  diameter  of  the  earth  is  7920  miles  and  that  of 
Mars  is  4230  miles,  find  the  ratio  of  their  surfaces. 

4.  If  the  diameter  of  the  moon  is  2160  miles,  find  the  ratio 
of  its  surface  to  that  of  the  earth. 

5.  The  volume  of  a  sphere  is  — - — •    If  V  represents  the 

o 

volume  of  a  sphere,  R  its  radius,  and  D  its  diameter,  show  that 
for  any  two  spheres,       y       j^s      j)s 

6.  The  diameter  of  the  sun  is  approximately  one  hundred 
and  nine  times  the  diameter  of  the  earth.  Find  the  ratio  of 
their  volumes. 

7.  From  Exercises  3  and  5  find  the  ratio  of  the  volumes  of 
Mars  and  the  earth. 

8.  Find  the  ratio  of  the  volumes  of  the  earth  and  the  moon. 

9.  The  areas  of  two  similar  triangles  are  to  each  other 
as  the  squares  of  any  two  corresponding  lines.  If  the  corre- 
sponding sides  of  two  similar  triangles  are  12  and  20  and 
the  area  of  the  first  is  90  square  inches,  find  the  area  of  the 
second. 

10.  The  areas  of  two  similar  triangles  are  147  and  300 
respectively.  If  the  base  of  the  first  is  10,  find  the  correspond- 
ing base  of  the  second. 

11.  If  ^jBC  is  any  triahgle  and  KR  is  a  line  parallel  to  BC, 
meeting  AB  at  ^-and  AC  at  R, 
then 

area  ABC  _  AB^  _  AC^  _  BC^ 
area  AKR  ~  AK^  ~  AR^  ~  KR^' 
If  in  the  accompanying  figure 
area    ABC  =  225  square  inches, 
area  AKR  =  81  square  inches,  and  AB  =  W  inches,  find  AK, 


242 


SECOND  COUESE  IN  ALGEBRA 


12.  In  the  figure  of  Exercise  11,  if  ABC  =  845  square  inches, 
BC  =  13  inc*hes,  and  KR  =  7  inches,  find  the  area  AKR. 

13.  If  in  the  figure  of  Exercise  11  triangle  AKR  equals  ^j 
of  the  trapezoid  KBCR  and  AC  =  15  inches,  find  AR  and  RC. 

14.  In  Exercise  13  substitute  ^  for  2^  and  solve  for  AR  to 
two  decimals. 

15.  If  in  the  figure  of  Exercise  11  the  triangle  is  equivalent 
to  the  trapezoid  and  AK  =  10,  find  KB  to  two  decimals. 

16.  A  certain  flagpole  casts  a  shadow  45  feet  long  at  the 
same  time  that  a  near-by  post  8  feet  high  casts  a  shadow 
4: J  feet  long.    Find  the  height  of  the  pole. 

17.  The  bisector  of  an  angle  of  a  triangle  divides  the  oppo- 
site side  into  segments  which  are  proportional  to  the  adjacent 
sides.  In  triangle  ABC  ii  AB  =  15,  BC  =  24:,  and  CA  =  25, 
find  the  segments  of  i^C  made  by  the  bisector  of  angle  A. 

\S.  If  a  plane  be  passed  par- 
allel to  the  base  of  a  pyramid 
(or  cone),  as  in  the  accompany- 
ing figure,  cutting  it  in  KRL, 
then  pyramid  D  —  ABC  :  pyra- 
mid D  -  KRL  =  DH^ :  DS^,  etc. 

If  in  the  adjacent  figure  the 
volumes  of  the  pyramids  are  8 
and  27  cubic  inches  respectively, 
and  the  altitude  DH  equals  15 
inches,  find  DS. 

19.  If  DH  in  the  accompany- 
ing figure  is  18  inches  and  the  volume  of  one  pyramid  is  one 
third  the  volume  of  the  other,  find  DS  to  two  decimals. 

20.  In  the  accompanying  figure  the  frustum  is  seven  eighths 
of  the  whole  pyramid,  (a)  If  DH  equals  16,  find  DS-,  (b)  if 
DH  =  X,  find  DS. 


RATIO,  PROPORTION,  AND  VARIATION      243 

21.  If  a  plane  parallel  to  the  base  divides  the  whole  pyra- 
mid into  two  parts  having  equal  volumes  and  DH  =  75,  find 
to  two  decimals  the  parts  into  which  the  plane  divides  DH. 

22.  The  volumes  of  two  similar  figures  are  to  each  other  as 
the  cubes  of  any  two  corresponding  edges.  Compare  the  vol- 
umes of  two  similar  solids,  the  edge  of  one  of  which  is  60% 
greater  than  the  corresponding  edge  of  the  other. 

23.  Compare  the  radii  of  two  spheres  whose  volumes  are 
to  each  other  as  125  :  27. 

133.  Variation.  The  word  quantity  denotes  anything 
which  is  measurable,  such  as  distance,  rate,  time,  and  area. 

Many  operations  and  problems  in  mathematics  deal  with  numeri- 
cal measures  of  quantities,  some  of  which  are  fixed  and  others  con- 
stantly changing.  Such  problems  as  deal  with  the  relation  of  the 
numerical  measures  of  at  least  two  changing  quantities  are  called 
problems  in  variation. 

The  theory  of  variation  is  really  involved  in  proportion. 
This  fact  will  become  evident  after  a  study  of  the  illustra- 
tions of  the  different  kinds  of  variation  here  given. 

The  equation  x='^y  may  refer  to  no  physical  quantities 
whatever,  yet  it  is  possible  to  imagine  y  as  taking  on  in 
succession  every  possible  numerical  value,  and  the  value 
of  X  as  changing  with  every  change  of  ?/,  and  consequently 
always  being  three  times  as  great  as  the  corresponding 
v^lue  of  y.  In  this  sense,  which  is  strictly  mathematical, 
X  and  y  are  variables. 

The  symbol  for  variation  is  oc,  and  x  cc  y  is  read  "  x 
varies  directly  as  ?/ "  or  ''  x  varies  as  ?/." 

134.  Direct  variation.  One  hundred  feet  of  copper  wire 
of  a  certain  size  weighs  32  pounds.  Obviously  a  piece  of 
the  same  kind  200  feet  long  would  weigh  64  pounds,  a 
piece  300  feet  long  would  weigh  96  pounds,  and  so  on. 


244  SECOND  COURSE  IN  ALGEBRA 

Here  we  have  two  variables,  W^  (weight)  and  L  (length), 
so  related  that  the  value  of  ^depends  on  the  value  of  X, 
and  in  such  a  way  that  W  increases  proportionately  as  L 
increases.  That  is,  W  is  directly  proportional,  or  merely  pro- 
portional, to  L,  Hence,  if  W^  and  W^  are  any  two  weights 
corresponding  to  the  lengths  L^  and  L<^  respectively, 

W,:  W,  =  L,:L,.  (1) 

The  fact  expressed  by  (1)  can  be  stated  in  the  form  of 
a  variation,  thus:  WccL. 

In  general,  if  x  cc  ^,  and  x  and  «/  denote  ani/  two  corre- 
sponding values  of  the  variables,  and  x-^  and  ?/j  a  particular 
pair  of  corresponding  values  of  these  variables, 

then  £  =  1.  (2) 

From  (2),  x  =  (^\^.  (3) 

X 

But  -^  is  a  constant,  being  the  quotient  of  two  definite 

numbers. 

Call  this  constant  K,  and  (3)  may  be  written 
x  =  Ky. 

That  is,  if  one  variable  varies  as  a  second,  the  first  always 
equals  the  second  multiplied  hy  some  co7istant. 

Thus  for  the  copper  wire  just  mentioned,  W=  j'VV  ^»  ^^  ^  ^• 
Here,  though  W  varies  as  L  varies,  W  is  always  equal  to  L  multi- 
plied by  the  constant  5%. 

EXERCISES 
1..  If  X  QC  y,  and  x  =  S  when  ^  =  8,  find  x  when  y  =12. 
Solution.    The  variation  is  direct.  ■" 

Therefore  -  =  — 

X      12 

Solving,  a;  =  4^. 


RATIO,  PEOPORTION,  AJ^D  VARIATION      245 

2.  li  x^y,  and  ic  =  8  when  y  =  15,  find  y  when  x  =  10. 

3.  If.  xcc  2/,  and  x  =  h  when  y  =  k,  find  y  when  x  =  r. 

4.  If  a?  oc  2/,  and  a?  =  2  when  y  =  5,  find  iT. 

135.  Inverse  variation.  If  a  tank  full  of  water  is  emptied 
in  24  minutes  through  a  smooth  outlet  in  which  the  area 
of  the  openmg  A  is,  1  square  inch,  an  outlet  in  which  A 
is  2  square  inches  would  empty  the  tank  in  one  half  the 
time,  or  in  12  minutes;  and  an  outlet  in  which  ^  is  3  square 
inches  would  empty  the  tank  in  8  minutes. 

Suppose  it  possible  to  increase  or  decrease  A  at  will. 
When  A  is  doubled  t  is  halved ;  when  A  is  trebled  t  is 
divided  by  3 ;  and  so  on.  We  then  have  in  t  (the  time 
required  to  empty  the  tank)  and  in  A  (the  area  of  the 
opening)  two  related  variables  such  that  if  A  increases,  t 
will  decrease  proportionally,  while  if  A  decreases,  t  will 
increase  proportionally. 

Now  let  t^  and  ^2  be  any  two  times  corresponding  to  the 
areas  A^  and  A^  respectively;  then 

t^:%  =  A^:A^.  (1) 

The  letters  and  the  subscripts  in  (1)  say :  The  first  time 
is  to  the  second  time  as  the  second  area  is  to  the  first  area. 

The    proportion    (1)    may   be   written    ^^  :  ^2  =  —  -  — 

A^     A^ 

where  the  subscripts  on  the  ^'s  and  those  on  the  ^'s  come 

in  the  same  order. 

First,  from  (1),  t^.  A^  =  t^.  A^.  •      (2) 

Dividing  (2)  by  ^1^2^     ^  =  ^'  (3) 

Whence  '<:];)  =  <i;)-  w 


246  SECOND  COURSE  IN  ALGEBRA 

Therefore  f ^ :  ^2  "=  —  •  —  (^) 

In  the  form  of  a  variation  (5)  becomes  ^  oc  — 

In  general,  x  varies  inversely  as  y  when  x  varies  as  the 
reciprocal  of  ?/ ;  that  is, 

.«i.  (6) 

And  if  X  and  7/  denote  any  two  corresponding  values 
of  the  variable,  and  x^  and  1;/^  a  particular  pair  of  corre- 
sponding values, 

xix-.  =  —  :  —  (7) 

Whence  —  =  ^,  or  ^«/  =  x-^y^.  (8) 

Vi    y 

But  x^y^  is  a  constant,  being  the  product  of  two  definite 
numbers.    Call  this  constant  K. 

Then  (8)  becomes  xy  =  K. 

That  is,  if  one  variable  varies  inversely  as  another,  the 
product  of  the  two  is  a  constant. 


EXERCISES 

1.  If  £c  varies  inversely  as  y,  and  x  =  S  when  y  =  5,  find  x 
when  y  =  15. 

Solution.    The  variation  is  inverse. 

Hence  8  :  a;  =  -  :  — r  • 

5    lo 

Solving,  a:  =  2§. 

2.  If  a;  QC  -5  and  x  =  1  wIkd  //  -  '-^').  fnid  x  when  ?/  =  10. 


RATIO,  PROPORTIOK,  AND  VARIATION      247 

3.  If  y  QC  -J  and  y  =  h  wlien  z  =  k,  find  y  when  z  =  r.    . 

4.  If  m  QC  -J  and  m  =  2  when  ??,  =  —  ?  find  m  when  n  =  12. 

5.  If  f  oc  -J  and  i^  =  2  when  tz,  =  8,  find  n  when  ?^  =  8. 

n 

6.  If  ?/J  QC  -?  and  w  =  100  when  d  =  4000,  find  w  when 
c^  =  5000. 

7.  If  ^  QC  -J  and  z^  =  4  when  r  =  25,  find  /C 

r 

8.  If  z^  oc  -,  and  w  =  200  when  d  =  4000,  find  ii:. 

cc 

136.  Joint  variation.  If  the  base  of  a  triangle  remains 
constant  while  the  altitude  varies,  the  area  will  vary  as 
the  altitude.  Similarly,  if  the  base  varies  while  the  alti- 
tude remains  constant,  the  area  w411  vary  as  the  base.  If 
both  base  and  altitude  vary,  the  area  varies  as  the  product 
of  the  two ;  that  is,  the  area  of  the  triangle  varies  jointly 
as  the  base  and  altitude.  Further,  if  at  any  time  A  denotes 
the  area  of  a  variable  triangle,  and  h^  and  5^  the  corre- 
sponding altitude  and  base,  then 

A  =  ^-  (1) 

If  A^  denotes  the  area  at  ani/  other  time,  and  h^  and  b^ 
the  corresponding  altitude  and  base,  then 

A  =  ^-  (2) 

Now  (1)^(2)  gives  A^:  A^  =  hj)^:  h^h^. 

In  the  form  of  a  variation  this  last  proportion  becomes 

Acchb. 


248  SECOND  COURSE  IN  ALGEBEA 

In  general,  any  variable  x  varies  jointly  as  two  others, 
y  and  2,  if  x<^yz',  (1) 

that  is,  if  X  varies  as  the  product  of  the  two. 

If  X  varies  jointly  as  y  and  2,  and  if  x^  y^  and  z  denote 
any  corresponding  values  of  the  variables,  while  x-^,  y^^  and 
z^  denote  a  particular  set  of  such  values,  then 

L=yL.  (2) 

From  (2),  x  =  i^^yz.  (3) 

But  in  (3)  the  fraction  — ^  is  a  constant,  since  o^j,  y^^ 

and  2j  are  particular  values  of  the  variables  x^  «/,  and  z. 
Calling  this  constant  JT,  we  may  write  xocyz  as  the 
equation  x=Kyz. 

One  variable  may  vary  directly  as  one  variable  (or  several  varia- 
bles) and  inversely  as  another  (or  several  others).  Also  one  variable 
may  vary  as  the  square,  or  the  cube,  or  the  square  root,  or  the  recip- 
rocal, or  as  any  algebraic  expression  whatever  involving  the  other 
variable  (or  variables). 

EXERCISES 

1.  If  £c  varies  jointly  as  y  and  z,  and  a?  =  24  when  2/  =  6  and 
^  =  8,  find  X  when  y  =  l^  and  ^  =  4. 

Solution.    The  variation  is  joint. 

24       6-8 
Therefore  —  =  -— — -  • 

X       18-4 

Solving,  X  =  36. 

2.  If  a?  varies  as  yz,  and  £c  =  10  when  y  =  15  and  z  =  6,  find 
X  when  y  —  ^  and  z  =  %. 

3.  If  yl  oc  hh,  and  A  =  30  when  A  =  5  and  ^^  =  12,  find  A 
when  h  —  1  and  5  =  10. 


RATIO,  PROPORTION,  AND  VARIATION      249 

4.  It*  A  oc  hb,  and  A  =  48  when  li  =  8  and  b  =  12,  find  /v. 

5.  li  X  varies  directly  as  y  and  inversely  as  z,  and  ic  =  10 
when  y  =  5  and  z  =  27,  find  x  when  2/  =  12  and  s  =  36. 

6.  If  F  varies  directly  as  T  and  inversely  as  P,  and  V  =  80 
when  P  =  30  and  T  =  300,  find  P  when  T  =  400  and  V  =  40. 

PROBLEMS 

1.  The  weight  of  any  object  below  the  surface  of  the  earth 
varies  directly  as  its  distance  from  the  center.  An  object 
weighs  172  pounds  at  the  surface  of  the  earth.  What  would 
be  its  weight  (a)  1000  miles  below  the  surface?  (b)  3000 
miles  below  the  surface  ?  (c)  at  the  center  of  the  earth  ? 
(Radius  of  the  earth  =  4000  miles.) 

2.  The  distance  which  sound  travels  varies  directly  as  the 
time.  A  soldier  measures  with  a  stop  watch  the  time  elapsing 
between  the  sight  of  the  flash  of  an  enemy's  gun  and  the 
sound  of  its  report.  If  sound  travels  1100  feet  per  second,  how 
far  off  was  the  enemy  Avhen  the  observed  time  was  5|  seconds  ? 

3.  When  the  volume  of  air  in  a  bicycle  pump  is  24  cubic 
inches,  the  pressure  on  the  handle  is  30  pounds.  Later  when 
the  volume  of  air  is  20  cubic  inches,  the  pressure  is  36  pounds. 
Assume  that  a  proportion  exists  here,  determine  whether  it  is 
direct  or  inverse,  and  find  the  volume  of  the  air  when  the 
pressure  is  42  pounds. 

4.  The  distance  (in  feet)  through  which  a  body  falls  from 
rest  varies  as  the  square  of  the  time  in  seconds.  If  a  body  falls 
16  feet  in  1  second,  how  far  will  it  fall  in  (a)  3  seconds  ? 
(b)  10  seconds? 

5.  The  intensity  (brightness)  of  light  varies  inversely  as 
the  square  of  the  distance  from  the  source  of  the  light. 
A  reader  holds  his  book  3  feet  from  a  lamp  and  later  6  feet 
distant.  At  which  distance  does  the  page  appear  the  brighter  ? 
How  many  times  as  bright  ? 

KK 


250  SECOND  COURSE  IN  ALGEBRA 

6.  A  lamp  shines  on  the  page  of  a  book  5.  feet  distant. 
Where  must  the  book  be  held  so  that  the  page  will  receive 
twice  as  much  light  ?  four  times  as  much  light  ? 

7.  The  area  of  a  circle  varies  as  the  square  of  its  radius 
The  area  of  a  certain  circle  is  154  square  inches  and  its  radius 
is  7  inches.  Find  the  radius  of  a  circle  whose  area  is  616  square 
inches. 

8.  The  area  illuminated  on  a  screen  by  a  spot  light  varies 
directly  as  the  square  of  the  distance  from  the  source  of 
the  light  to  the  screen.  If  the  lighted  area  at  a  distance  of 
40:  feet  is  a  circle  of  diameter  10  feet,  find  the  diameter  of  the 
illuminated  circle  at  a  distance  of  15  feet. 

9.  The  weight  of  an  object  above  the  surface  of  the  earth 
varies  inversely  as  the  square  of  its  distance  from  the  center 
of  the  earth.  An  object  weighs  172  pounds  at  the  surface  of 
the  earth.  What  would  it  weigh  (ci)  1000  miles  above  the 
surface  ?  (h)  3000  miles  above  the  surface  ?  (<•)  5000  miles 
above  the  surface  ? 

10.  How  far  above  the  surface  of  the  earth  would  a  150-pound 
object  have  to  be  placed  so  that  its  weight  would  be  reduced 
one  third  ? 

11.  The  weight  of  a  sphere  of  given  material  varies  directly 
as  the  cube  of  its  radius.  Two  spheres  of  the  same  material 
have  radii  3  inches  and  5  inches  respectively.  The  first  weighs 
8  pounds.    Find  the  weight  of  the  second. 

12.  The  time  required  by  a  pendulum  to  make  one  vibra- 
tion varies  directly  as  the  square  root  of  its  length.  If  a 
pendulum  100  centimeters  long  vibrates  once  in  1  second, 
find  the  time  of  one  vibration  of  a  pendulum  81  centimeters 
long. 

13.  Find  the  length  of  a  pendulum  which  vibrates  once  in 
2  seconds :  once  in  7  seconds. 


RATIO,  PROPORTION,  AND  VARIATION      251 

14.  The  pressure  of  wind  on  a  plane  surface  varies  jointly 
as  the  area  of  the  surface  and  the  square  of  the  wind's  velocity. 
The  pressure  on  1  square  foot  is  .9  pound  when  the  rate  of 
the  wind  is  15  miles  per  hour.  Find  the  velocity  of  the  wind 
when  the  pressure  on  1  square  yard  is  72.9  pounds. 

15.  The  pressure  of  water  on  the  bottom  of  a  containing 
vessel  varies  jointly  with  the  area  of  the  bottom  and  the  depth 
of  the  water.   When  the  water  is  1  foot  deep  the  pressure  on 

1  square  foot  of  the  bottom  is  62.5  pounds.  Find  the  pressure 
on  the  bottom  of  a  circular  tank  of  14  feet  diameter  in  whicdi 
the  water  is  10  feet  deep. 

16.  The  cost  of  ties  for  a  railroad  varies  directly  as  the 
length  of  the  road  and  inversely  as  the  distance  between  the 
ties.    The  cost  of  ties  for  a  certain  piece  of  road,  the  ties  being 

2  feet  apart,  was  $1320.  Find  the  cost  of  ties  for  a  piece  twenty 
times  as  long  as  the  first  if  tlie  ties  are  2^  feet  apart. 


CHAPTER  XIX 

LOGARITHMS  , 

137.  Introduction.  Logarithms  were  invented  to  shorten 
the  work  of  extended  numerical  computations  which  involve 
one  or  more  of  the  operations  of  multiplication,  division, 
involution,  and  evolution.  Their  use  has  decreased  the 
labor  of  computing  to  such  an  extent  that  many  calcula- 
tions which  would  require  hours  without  the  use  of  loga- 
rithms can  be  performed  with  their  aid  in  one  tenth  of 
the  time  or  less. 

138.  Definition  of  logarithm  and  base.  If  we  write  the 
equation  ^^^y^  ^^^ 

we  express  therein  the  essential  relation  between  a  number, 
n,  and  its  logarithm,  /,  for  a  given  base,  h.  In  the  notation 
of  logarithms  this  is  written 

logt«  =  ?,  (2) 

and  it  is  read  ''the  logarithm  oi'n  to  the  base  h  equals  /." 
We  can  define  verbally  in  one  statement  both  logarithm 
and  base  as  follows: 

The  logarithm  of  a  given  mimher  is  the  exponent  in  the 
power  to  which  ariother  number^  called  the  base^  7nu8t  be 
raised  in  order  to  equal  the  given  number. 

It  is  important  to  realize  that  equations  (1)  and  (2)  are 
merely  two  different  ways  of  expressing  precisely  the  same 
relations,  one  the  exponential  way,  the  otlier  the  logarithmic^ 

262 


LOGARITHMS  253 

Above  all  it  is  necessary  to  keep  in  mind  the  fact  that  a 
logarithm  is  an  exponent. 

Thus  in  32  =  2^  the  given  number  is  32,  the  base  is  2,  and  tlie 
logarithm  is  5  ;  that  is,  log2  32  =  5. 

139.  Systems  of  logarithms.  The  base  of  the  common^  or 
Brifigs,  system  of  logarithms  is  10.  Hence  a  table  of  common 
logarithms  is  really  a  table  of  exponents  of  the  number  10. 
Since  the  greater  portion  of  these  exponents  are  approxi- 
mate values  of  irrational  numbers,  it  follows  that  compu- 
tations by  means  of  logarithms  give  only  approximate 
results.  Tables  exist,  however,  in  which  each  logarithm  is 
given  to  twenty  or  more  decimals ;  hence  practically  any 
desired  degree  of  accuracy  can  be  obtained  by  using  the 
proper  table.  This  system  is  used  in  numerical  work  almost 
exclusively.  The  table  on  pages  266-267  is  a  table  of 
common  logarithms  carried  to  four  decimal  places. 

The  only  other  system  of  logarithms  used  in  computa- 
tions is  called  the  natural  system.  It  has  for  its  base  the 
irrational  number  2.7182 -f,  which  is  usually  denoted  by 
the  letter  e  and  is  used  mainly  for  theoretical  purposes. 
-  It  can  be  proved  that  the  laws  given  on  pages  91-92, 
governmg  the  use  of  rational  exponents,  hold  for  irrational 
exponents.  In  the  work  on  logarithms  this  fact  will  be 
assumed. 

ORAL  EXERCISES    ' 

1.  If  8  =  2\  x  =  ?  log^S  =  ? 

2.  If  1000  =  10^«  =  ?logj^  1000  =  ? 

3.  legale  ==  ?  log3  81  =  ?  log^625  =  ? 

4.  If  b^=U,h  =  ?    If  h^  =  243,  b  =  ? 

5.  10^  =  ?  logj^lO  =  ?  8.  3«  =  ?  loggl  =  ? 

6.  lO'^  =  ?  log.olOO  =  ?  9.  5«  =  ?  loggl  =  ? 

7.  W  =  ?  log^^l  =  ?  10.  n'=?  log;,l=  ? 


25^ 


SECOND  COUKSE  IN  ALGEBRA 


Number 

Base 

Logarithm 

Number 

Base 

Logarithm 

11. 

8 

2 

? 

24. 

? 

4 

3 

12. 

32 

2 

? 

25. 

V 

6 

2 

13. 

64 

4' 

o 

26. 

9 

V 

J 

14. 

125 

5 

V 

27. 

8 

•/ 

3 
2 

15. 

1000 

10 

? 

28. 

16 

? 

3 

16. 

16 

' 

? 

2 

29. 

2 

4 

? 

17. 

81 

? 

2 

30. 

3 

9 

? 

18. 

81 

? 

4 

31. 

2 

8 

? 

19. 

40 

f 

2 

32. 

3 

81 

? 

20. 

216 

f 

3 

33. 

T^ 

49 

<> 

21. 

V 

10 

3 

34. 

8 

■1. 

•/ 

22. 

? 

10 

»> 

35. 

4 

8 

V 

23. 

? 

1( 

3 

1 

36. 

.1 

10 

•> 

liead  in  tlic  notation  of  lojjiiritlinis : 

37.  300  =10-^*^  42.  1730  =  103-*^. 

38.  65:::=10^«^  43.  173  =  W--^«^ 

39.  4=:10"".  44.  1.73  =  10-'«. 

40.  1  =  10^  45.  .173  =  10-1  + •-'^\ 

41.  .10  =  10-^        •  46.  .0173  =  10- '-  +  •-««. 

Read  Exercises  47-49  and  54-56  as  powers  of  10 : 

47.  log  3  =  .48.        48.  log  20  =  1.301.        49.  log  4.9  =  .69. 

50.  log^^lOO  +  log^^lOOO  4-  log,olO,000  =  ? 

51.  log^,10  +  log^^.Ol  -  logj^l  =  ?      54.  log  490  =  2.69. 

52.  log^S  +  log327  4-  log,l  =  ?  55.  log  .0049  =  -  3  +  .69. 
.  53.  log39  +  log^64  =  ?  56.  log  381  =  2.58. 


LOGARITHMS  255 

Biographical  Note.  John  Napier.  Although  many  scientists  have 
been  honored  with  titles  on  account  of  their  discoveries,  very  few  of  the 
titled  aristocracy  have  become  distinguished  for  their  mathematical 
achievements.  A  notable  exception  to  this  rule  is  found  in  John  Napier, 
Lord  of  Merchiston  (1550-1617),  who  devoted  most  of  his  life  to  the 
problem  of  simplifying  arithmetical  operations.  Napier  was  a  man  of 
wide  intellectual  interests  and  great  activity.  In  connection  with  the 
management  of  his  estate  he  applied  himself  most  seriously  to  the  study 
of  agriculture,  and  experimented  with  various  kinds  of  fertilizers  in  a 
somewhat  scientific  manner,  in  order  to  find  the  most  effective  means  of 
reclaiming  soil.  He  spent  several  years  in  theological  writing.  When  the 
danger  of  an  invasion  by  Philip  of  Spain  was  imminent  he  invented 
several  devices  of  war.  Among  these  were  powerful  burning  mirrors, 
and  a  sort  of  round  musket-proof  chariot,  the  motion  of  which  was  con- 
trolled by  those  within,  and  from  which  guns  could  be  discharged  through 
little  portholes.  " 

But  by  far.  the  most  serious  activity  of  Napier's  life  was  the  effort  to 
shorten  the  more  tedious  arithmetical  processes.  He  invented  the  first 
approximation  to  a  computing  machine,  and  also  devised  a  set  of  rods, 
often  called  Napier's  bones,  which  were  of  assistance  in  multiplication. 
His  crowning  achievement,  however,  was  the  invention  of  logarithms,  to 
which  he  devoted  fully  twenty  years  of  his  life. 

140.  Steps  preceding  computation.  Before  computation 
by  means  of  the  table  can  be  taken  up,  two  processes 
requiring  considerable  explanation  and  practice  must  be 
mastered. 

/.   To  find  from  the  table  the  logarithm  of  .a  given  iiumher. 
II.   To  find  from  the  table  the  number  corresponding  to  a 
given  logarithm, 

141.  Characteristic  and  mantissa.  Unless  a  number  is 
an  exact  power  of  10,  its  logarithm  consists  of  an  integer 
and  a  decimal. 

This  fact  is  illustrated  in  Exercises  37-46,  p.  254. 
The  integral  part  of  a  logarithm  is  called  its  characteristic. 
The  decimal  part  of  a  logarithm  is  called  its  mantissa. 
Log  200  =  2.301.    Here  2  is  the  characteristic  and  .301 
is  the  mantissa. 


256  SECOND  COURSE  IN  ALGEBRA 

The  characteristic  of  any  number  is  obtained  not  from 
a  table  of  logarithms  but  by  an  inspection  of  the  number 
itself,  according  to  rules  which  will  now  be  derived. 

104  =  10,000;  that  is,  the  log  10,000  =  4. 


103  =  1000; 

that  is,  the  log  1000 

=  8. 

102  =  100; 

that  is,  tlie  log  100 

=  2. 

101=10; 

that  is,  the  log  10 

=  1. 

100  =  1; 

that  is,  the  log  1 

=  0. 

10-i=.l; 

that  is,  the  log  .1 

=  -1. 

10-2  =  .01; 

that  is,  the  log  .01 

=  —2. 

10-3  =.001; 

that  is,  the  log  .001 

=  -3. 

The  preceding  table  indicates  between  what  two  integers 
the  logarithm  of  a  number  less  than  10,000  lies.  This 
determines  the  characteristic. 

Since  542  lies  between  100  and  1000  (that  is,  between 
102  and  103),  log  542  must  lie  between  2  and  3  and  must 
equal  2  (characteristic)  plus  a  decimal  (mantissa). 

And  since  .0045  lies  between  .001  and  .01  (that  is, 
between  10"^  and  10"'^),  log  .0045  =  — 3  plus  a  positive 
decimal  or  —  2  plus  a  negative  decimal. 

For  the  determination  of  the  characteristic  of  a  positive 
number  we  have  the  following  rules: 

/.  The  characteristic  of  a  number  greater  than  1  is  one  less 
than  the  number  of  digits  to  ike  left  of  the  decimal  point, 

II.  The  characteristic  of  a  number  less  than  1  is  negative 
and  numerically  one  greater  than  the  number  of  zeros  between 
the  decimal  point  and  the  first  significant  jigwre. 

Accordingly  the  characteristic  of  2536  is  3;  of  6  is  0; 
of  .4  is  -1;  of  .032  is  -  2 ;  of  .00086  is  -4. 


JOHN  NAPIER 


LOGARITHMS 

25 

ORAL 

EXERCISES 

What  is 

the  characteristic 

of  the  following : 

1.  347. 

5.  35. 

9.  97.2. 

13. 

.00972. 

2.  9864. 

6.  972. 

10.  9.72. 

14. 

30.467. 

3.  95. 

7.  9720. 

11.  .972. 

15. 

.5000. 

4.  7. 

8.  97200. 

12.  .0972. 

16. 

.000375. 

The  table  on  pages  2G6-267  gives  the  mantissas  of 
numbers  from  10  to  999.  Before  each  mantissa  a  decimal 
point  is  understood. 

The  numbers  5420,  542,  5.42,  .0542,  and  .000542  are 
spoken  of  as  composed  of  the  same  significant  digits  in  the 
same  order.  They  differ  only  m  the  position  of  the  deci- 
mal pohit,  and  consequently  their  logarithms  to  the  base  10 
will  have  different  characteristics,  but  they  Avill  have  the 
same  mantissa. 

The  last  two  points  are  easily  illustrated  b}^  any  two 
numbers  Avhich  have  the  same  significant  digits  in  the 
same  order. 

log  5.42  =  .734,  or  5.42  =  10-'34; 

5.42  .  102 _  542  =  10-1 .  102 _  102.734. 

Therefore  log  542  =  2. 734. 

The  property  just  explamed  does  not  belong  to  a  system 
of  logarithms  in  which  the  base  is  any  number  other  than  10. 
Thus,  if  the  base  is  100,  the  most  convenient  number  after 
10,  the  logarithms  of  5420,  542,  54.2,  and  5.42  are  respec- 
tively 1.8670,  1.3670,  .8670,  and  .3670.  While  a  certain 
regularity  in  characteristic  and  mantissa  can  be  seen  here, 
it  is  obvious  that  the  rules  for  obtaining  them  would  not 
be  so  simple  as  they  are  for  the  base' 10.  Moreover,  it  can 
be  seen  that  tables  of  a  given  accuracy  are  far  shorter  when 
the  base  is  10  than  they  would  be  with  any  other  base. 


258  SECOND  COURSE  IN  ALGEBRA 

142.  Use  of  the  table.  To  obtain  the  logaritlim  of  a 
imiiiber  of  three  or  fewer  significant  figures  from  tlie 
table,  we  have  the 

Rule.    Determine  the  characteristic  hy  inspection. 

Find  in  colmnn  N  the  first  two  sif/nificant  figures  of  the 
given  7iimiber.  In  the  roiv  with  these  and  iii  the  column  headed 
hy  the  third  figure  of  the  number,  find  the  required  mantissa. 

ORAL  EXERCISES 
Find  tlie  logarithm  of  the  following : 

1.  263.  4.  56.  7.  3.7.  10.  7. 

2.  375.  5.  560.  8.   3700.  11.  932. 

3.  729.  6.  37.  9.  5.  12.  .932. 

Solution.  The  characteristic  of  .932  is  —  1  and  the  mantissa  is 
.9694.  Hence  log  .932  =  —  1  +  .9694.  This  is  usually  written  in  the 
abhreviated  form,  1.9694.  The  mantissa  is  always  kept  positive  in 
order  to  avoid  the  addition  and  subtraction  of  both  positive  and 
negative  decimals,  which  in  ordinary  practice  contain  from  three  to 
five  figures.  Negative  characteristics,  being  integers,  are  compara- 
tively easy  to  take  care  of.  (The  student  should  note  that  log  .932 
is  really  negative,  being  —1-1-  .9694,  or  —  .0306.) 

13.  .563.  15.  .00376.  17.  .0202.  19.  3.86. 

14.  .0637.  16.  .00468.  18.  725000.  20.  .987. 

143.  Interpolation.  The  process  of  finding  the  logarithm 
of  a  number  not  found  in  the  table,  from  the  logarithms 
of  two  numbers  which  are  found  there,  or  the  reverse  of 
this  process,  is  called  interpolation. 

If  we  desire  the  logarithm  of  a  number  not  in  the  table, 
say  7635,  we  know  that  it  lies  between  the  logarithms  of 
7630  and  7640,  which  are  given  in  the  table.  Since  7635 
is  halfway  between  7630  and  7640,  we  assume,  though  it 
is  not  strictly  true,  that  the  required  logarithm  is  halfway 


LOGAllITHMS  259 

between  their  logarithms,  3.8825  and  3.8831.  In  order  to 
find  log  7635  we  first  look  up  log  7630  and  log  7640  and 
then  take  half  (or  .5)  their  difference  (this  difference 
may  usually  be  taken  from  the  column  headed  D)  and 
add  it  to  log  7630.     This  gives 

log  7635  =  3.8825 +.5  x.0006  =  3.8828.    . 

Were,  we  finding  log  7638,  we  should  take  .8  of  the 
difference  between  log  7630  and  log  7640  and  add  it  to 
log  7630  as  follows : 

log  7638  =  3.8825 +.8  X  .0006 
=  3.8825 +.00048 
=  3.8825 +.0005 

=  3.8830. 

Observe  that  in  using  four-place  tables  one  should  not 
carry  results  to  five  figures.  If  the  fifth  figure  is  5,  6,  7, 
8,  or  9,  omit  it  and  increase  the  fourth  figure  by  1 ;  that 
is,  obtain  results  to  the  nearest  figure  in  the  fourth  place. 

For  finding  the  logarithm  of  a  number  we  have  the 

Rule.  Prefix  the  proper  characteristic  to  the  mantissa  of  the 
first  three  significant  figures. 

Then  multiply  the  difference  hettveen  this  mantissa  and  the 
next  greater  mantissa  in  the  table  (^called  the  tabular  differ- 
ence') by  the  remaining  figures  of  the  number  preceded  by  a 
decimal  point. 

Add  the  product  to  the  logarithm  of  the  first  three  figures^ 
taking  the  nearest  decimal  in  the  fourth  place. 

In  this  method  of  interpolation  we  have  assumed  that 
the  increase  in  the  logarithm  is  directly  proportional  to 
the  increase  in  the  number.  As  has  been  said,  this  is 
not  strictly  true,  yet  the  results  here  obtained  are  nearly 
always  correct  to  the  fourth  decimal  place. 


260  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 
Find  the  logaritliin  of  the  following : 

1.  3625.  5.  646.8.  9.  705.50.  13.  30.07. 

'     2.  464.7.  6.  82.543.         10.  3.0075.  •  14.  3.1416. 

3.  52.73.  7.  10.101.         11.  .00286.  15.  2.71828. 

4.  42.75.  8.  500.35.         12.  .0007777.  16.  .0023456. 

144.  Antilogarithms.  An  antilogarithm  is  the  number  cor- 
responding to  a  given  logarithm.  Thus  antilog  2  equals  100. 

If  we  desire  the  antilogarithm  of  a  given  logarithm, 
say  4.7308,  we  proceed  as  follows:  The  mantissa  .7808  is 
found  in  the  row  which  has  53  in  column  N  and  in  the 
column  which  has  8  at  the  top.  Hence  the  first  three  sig- 
nificant figures  of  the  antilogarithm  are  538.  Since  the 
characteristic  is  4,  the  number  must  have  five  digits  to  the 
left  of  the  decimal  pohit.    Thus  antilog  4.7308  =  53,800. 

Therefore,  if  the  mantissa  of  a  given  logarithm  is  found 
in  tlie  table,  its  antilogarithm  is  obtained  by  tlie 

Rule.  Find  the  roiv  and  the  colmnn  in  which  the  given 
mantissa  lies.  In  the  row  found  take  the  two  figures  which 
are  in  column  N  for  the  first  two  significant  figures  of  the 
antilogarithm  and  for  the  third  figure  the  number  at  the  top 
of  the  column  in  tvhich  the  mantissa  stands. 

Place  the  decimal  point  as  ind/'cafcd  li/  ihc  rltaractenstic. 

ORAL  EXERCISES 
Find  the  antilogarithm  of  the  following : 

1.  3.8768.               6.  7.5866-10.  10.  4.1335. 

2.  1.8035.                   Hint.   7. 58«0  -  10  =  8.58(5(5.  11.    6.9154. 

3.  .5763.                  7.9.2455-10.  12.8.3464. 

4.  1.3747.               8.  4.1335  - 10.  13.  0.8882. 

5.  2.7649.                9.  5.7875  -  (>.  14.  5.9689. 


LOGARITHMS  261 

If  the  mantissa  of  a  given  logarithm,  as  1.5271,  is  not 
in  the  table,  the  antilogarithm  is  obtained  by  interpolation 
as  follows: 

The  mantissa  5271  lies  between 

.5263,  the  mantissa  of  the  sequence  336, 
and         .5276,  the  mantissa  of  the  sequence  337. 

Therefore  the  antilogarithm  of  1.5271  lies  between  33.6 
and  33.7.  Since  the  tabular  difference  is  13  and  the  dif- 
ference between  .5263  and  .5271  is  8,  the  mantissa  .5271 
lies  -^^  of  the  way  from  .5263  to  .5276.  Therefore  the  re- 
quired antilogarithm  lies  ^^  of  the  way  from  33.6  to  33.7. 

Then  antilog  1.5271  =  33.6  +  ^83  x  .1,      • 

and  33.6 +.061=  33.66. 

Therefore  when  the  mantissa  is  not  found  in  the  table 
we  have  the 

Rule.  Write  the  number  of  three  figures  corresponding  to  the 
lesser  of  two  mantissas  hetiveen  which  the  given  mantissa  lies. 

Subtract  the  less  mantissa,  from  the  given  one  and  divide 
the  remainder  by  the  tabular  difference  to  two  decimal  places. 
If  the  second  digit  is  5  or  more.,  increase  the  first  digit  by  1; 
if  less  than  5.,  omit  it. 

Annex  the  resulting  digit  to  the  three  already  found  and 
place  the  decimal  point  where  indicated  by  the  characteristic. 

EXERCISES 
Find  the  antilogarithms  of  the  following : 

1.  1.5523.                 5.  1.2566.  9.  9.2664  - 10. 

2.  2.3821.                 6.  7.3572  -  10.  10.  .7729. 

3.  0.6790.                 7.  9.8327  -  10.  11.  7.1060  -  10. 

4.  2.5720.                 8.  5.9613  -  8.  12.  6.2318  -:-  10. 


262 


SECOND  COURSE  IN  ALGEBRA 


145.   Multiplication.     Multiplication   by  logarithms  de- 
pends on  the 

Theorem.    The  logarithm  of  the  product  of  tivo  numbers 
is  the  sum  of  the  logarithms  of  the  numhers. 

.  That  is,  for  the  numbers  a  and  x 

\og,,{a  .  X)  =  log^a  -h  log^a-. 

Proof.    Let  log^a  =  l^,  (1) 

and  log/^x  =  h.  (2) 

From  (i),  a  =  bh.  (3) 

From  (2),  x  =  hk.  (4) 

(3)  X  (4),  ax  =  ¥1  +  12.  (5) 

Therefore  lt)g^«.r  =  li-^  U 

=  logf,a  +  \ogf,x. 


Solution. 


EXERCISES 

Perform  the  indicated  operation  by  logarithms 

1.  18  X  25. 

log  18  =1.2553 
log25  =  1.3979 
log  (18  X  25)  =  2.G532  (adding) 
antilog  2.6532  =  450. 

6.  386  X  27. 

7.  432  X  263. 

8.  589  X  375. 

9.  4326  X  497. 


2.  37  X  28. 

3.  29  X  9. 

4.  9.8  X  6. 

5.  42  X  3.3. 


10.  2870  X  3754. 

11.  286.7  X  2.391. 

12.  3.412  X  2.526. 

13.  432  X  .574. 


Solution. 


log  432  =  2.6355  =    2.6355 
log  .574  =  1.7589  =    9.7589 


10 


log (432  X  .574)  =  2.3944  =  12.3944  -  10  (adding) 
autilog  2.3944  =  247.9. 


LOGARITHMS  263 

Since  the  mantissa  is  always  positive,  any  number  carried  over  from 
the  tenths'  cohimn  to  the  units'  column  is  positive.  This  occurs  in 
the  preceding  solution,  where  .6  + -7  =  1.3,  giving  +  1  to  be  added 
to  the  sum  of  the  characteristics  +  2  and  —  1,  in  the  units'  column. 
Mistakes  in  such  cases  will  be  few  if  the  logarithms  with  negative 
characteristics  be  written  as  in  the  9—10  notation  on  the  right. 

In  the  preceding  example  and  in  others  which  follow,  two  methods 
are  given  for  writing  the  logarithms  which  have  negative  character- 
istics. This  is  done  to  illustrate  those  cases  in  which  the  second  of 
th<5^two  ways  is  preferable.  It  should  be  understood  that  in  practice 
one,  but  not  necessarily  both,  of  these  methods  is  to  be  used. 

14.  385  X  .647.  19.  .6381  x -.01897. 

15.  571  X    073  Hint.  Determine  by  inspection  the 

sign  of  the  product.   Then  operate  as 

16.  37.6  X  .00865.  if  all  signs  were  positive. 

17.  .0476  X  673.  20.  675  x  -  .0286. 

18.  .07325  X  6.354.  21.  -.437  x  -.0046. 

146.  Division.    Division  by  logarithms  depends  on  the 
Theorem.   The  logarithm   of  the  quotient  of  two  numbers 

is  the  logarithm  of  the  dividend  minus  the  logarithm  of  the 

divisor. 

That  is,  for  the  numbers  a  and  x 


logfc 

a 

'  X 

=  logf^a  -  log^.r. 

Proof.  Let 
i 

log^a  =  /j, 

(1) 
(2) 

From  (1), 

a  =  Uu 

(•^) 

From  (2), 

X  =  hh. 

(4) 

(3) -^(4), 

X 

Therefore 

=  logta  -  logftA-. 

264  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 
Divide,  iising  logarithms : 
1.  891^27. 

Solution.  log  891  =  2.9499 

log  27  =1.4314 
log  (891  -^  27)  =  1.5185  (subtracting) 
antilog  1.5185  =  33. 

2.  96-5-32.  5.  439-- 27.1.  8.  9896 -- 52.78. 

3.  888  -T-  47.  6.  3860  -^  4.32.  9.  6732  -f-  7.81. 

4.  976^361.  7.  4627^281.  10.  3.26 -- .0482. 

Solution,  log  3.26  =  0.5132  =  10.5132  -  10 

log. 0482  =  2.6830=8^6830-10 


log  (3.26  -i-  .0482)  =  1.8302  =  1.8302  -  0 
antilog  1.8302  =  67.64. 

11.  2.35-^.0683.  _    347  x  (-  625) 

12.  4.86 --.751.  *            346 

13.  .0635-^.277.  473.2  x  4.78 

14.  .2674 --3.66.  *        -68.3 

15.  .07882  H- 68.72.  9.63  x  -0892 

16.  356  X  392-- 128.  ^^'        .00635 

147.  Involution.  Involution  by  logarithms  depends  on  the 

Theorem.     The  logarithm  of  the  rth  power  of  a  number  is 
r  times  the  logarithm  of  the  number. 

That  is,  for  the  numbers  r  and  x,  log^a^  =  r  log^a;. 
Proof.   Let  logftx  =  I.  (1) 

Then  x  =  ¥.  (2) 

Raising  both  members  of  (2)  to  the  rth  power, 

xr  =  bri. 
Therefore  logta:^  =  rl 

=  r  log&a:. 


LOGARITHMS  265 

EXERCISES 

Compute,  using  logarithms : 

1.  (2.73)«. 

Solution.  log  2.73  =  .4362. 

log  (2.73)3  =  1.3086  (multiplying  by  3). 
antilog  1.3086  =  20.33. 

2.  (6.32)*.  3.  (34.26)1  4.  (6.715)1 

5.  (.425)1 

Solution.  log  .425  =  1.0284  =  9.0284  -  10. 

log  (.425)3  =  2.8852  =  28.8852  -  30. 
antilog  2.8852  =  .07677. 

6.  (.362)1  '  9.  (486.2)2 .  (3.85)3. 

7.  (.0972)1  10.  (.375)^  •  (62.5)1 

8.  (.003597)^  11.  (^.25)^  -  (1.232)3. 

148.  Evolution.    Evolution  by  means  of  logarithms  de- 
pends on  the 

Theorem.     The  logarithm  of  the  real  rth  root  of  a  number 
is  the  logarithm  of  the  number  divided  hy  r. 

That  is,  for  the  real  numbers  r  and  w,  logf^^n=  —  logjn. 

Proof.    Let  log,/*  =  f-  (1) 

Then  n  =  bK  (2) 

Extracting  the  rth  root  of  both  members  of  (2), 

1  11 

(ny  =  (h^Y  =  hr,  (3) 

1        7       log,  n 
Therefore  log.  (nV  =  -  =  -^^^^  •  (4) 

^         r  r 

RE 


266 


SECOND  COURSE  IN  ALGEBRA 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

42 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

38 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

35 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

32 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

30 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

28 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

20 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

25 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

24 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

22 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

20 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3641 

3560 

3579 

3598 

19 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

18 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

18 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

17 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4266 

4281 

4298 

16 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4'425 

4440 

4456 

16 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

16 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

15 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

14 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

6024 

5038 

14 

32 

5051 

5065 

6079 

6092 

5105 

5119 

5132 

5145 

6159 

6172 

13 

33 

5185 

5198 

5211 

6224 

5237 

5250 

5263 

5276 

6289 

6302 

13 

34 

5315 

5328 

6340 

5353 

6366 

6378 

6391 

5403 

5416 

5428 

13 

35 

5441 

5453 

6465 

6478 

6490 

5502 

6514 

5527 

6639 

5551 

12 

36 

5563 

5575 

5587 

6599 

5611 

5623 

5635 

6647 

5658 

5670 

12 

37 

5682 

5694 

5705 

6717 

6729 

5740 

5752 

5763 

6775 

5786 

12 

38 

5798 

5809 

5821 

6832 

6843 

6855 

6866 

6877 

6888 

5899 

11 

39 

5911 

5922 

5933 

6944 

6955 

6966 

6977 

6988 

6999 

6010 

11 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

11 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

10 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

10 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

10 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

10 

45 

6532 

6542 

6551 

6561 

6571 

6680 

6590 

6599 

6609 

6618 

10 

46 

6(J28 

6637 

6646 

6666 

666.5 

6676 

6684 

6693 

6702 

6712 

9 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

9 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

9 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6965 

6964 

6972 

6981 

9 

60 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

9 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

8 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

8 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

8 

64 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

8 

LOGARITHMS 


267 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

8 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

8 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

8 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

6 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

6 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

6 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

6 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

6 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

6 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

6 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

6 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

6 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

6 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

6 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

6 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

5 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

5 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

5 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

5 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

5 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

5 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

5 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

5 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

5 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

5 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

5 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

5 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

5 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

5 

93 

9685 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

5 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

5 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

5 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

5 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

4 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

4 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

4 

268  SECOND  COURSE  IN  ALGEBKA 


EXERCISES 

Compute,  using 

logarithms : 

1.  -y/sje. 

Solution. 

log  376  =  2.5752. 

log  V  376  =  .8584  (dividing  by  3). 

Then 

antilog  .8584  =  7.218. 

2.   ^783 

3.    S/1435.             4.   ^3421 

5.    A/.000639. 

Solution. 

log  .000639  =  4.8055. 

If  one  divided  4.8055  as  it  stands  by  3,  he  would  be  likely  to 
confuse  the  negative  characteristic  and  the  positive  mantissa.  This 
and  other  difficulties  may  easily  be  avoided  by  adding  to  the  char- 
acteristic and  subtracting  from  the  resulting  logarithm  any  integral 
multiple  of  the  index  of  the  root  which  will  make  the  characteristic 
positive. 

Thus  log  .000639  =  2.8055  -  6. 

iog\/.000639  =  .9:^52-2  (dividing  by  3). 
Then  antilog  2.9352  =  .08614. 

6.  VX)756.        11.  (-  6.387)^.  14.   a/269. 


7.^.0007624  ,^^3-^^      15.  v/i74^. 


t:     12.   - 


8.  V.005679.  M    (3.423)« 

^-  ^^^-^^  •  1(43.56)^.7.984 

10.  (4.925)'.  N       (7.623)«       '  17.  ^^V?  ^-07241. 

18.  Determine  the  logarithms  of  5732,  573.2,  57.32,  and 
5.732  to  the  base  10  and  to  the  base  100.  Compare  the  results. 
What  fact  about  logarithms  do  these  results  emphasize  ? 


LOGARITHMS  269 

Note.  The  preceding  four-place  table  will  usually  give  results 
correct  to  one  half  of  one  per  cent.  Five-place  tables  give  the  man- 
tissa to  five  decimal  places  of  the  numbers  from  1  to  9999  and,  by 
interpolation,  the  mantissa  of  numbers  from  1  to  99,999.  Such  tables 
give  results  correct  to  one  twentieth  of  one  j)er  cent,  a  degree  of 
accuracy  which  is  sufficient  for  most  engineering  work. 

Six-place  tables  give  the  mantissa  to  six  decimals  for  the  same 
range  of  numbers  as  a  five-place  table,  but  the  labor  of  using  a  six- 
place  table  is  much  greater  than  that  of  using  a  five-place  one. 

Seven-place  tables  contain  the  mantissas  of  the  numbers  from 
1  to  99,999.  Such  tables  are  needed  in  certain  kinds  of  engineering 
work  and  are  of  constant  use  in  astronomy. 

In  place  of  a  table  of  logarithms  engineers  often  use  an  instru- 
ment called  a  slide  rule.  This  is  really  a  mechanical  table  of 
logarithms  arranged  ingeniously  for  rapid  practical  use.  Results  can 
be  obtained  with  such  an  instrument  far  more  quickly  than  with  an 
ordinary  table  of  logarithms,  and  that  without  recording  or  even 
thinking  of  a  single  logarithuL.  A  slide  rule  ten  inches  long  usually 
gives  results  correct  to  three  figures.  In  work  requiring  greater 
accuracy  a  larger  and  more  elaborate  instrument  which  gives  a  five- 
figure  accuracy  is  used. 

149.  Exponential  equations.  An  exponential  equation  is  an 
equation  in  which  the  unknown  occurs  in  an  exponent. 

Many  exponential  equations  are  readily  solved  by  means 
of  logarithms,  since  log  a^  =x  log  a. 

Thus  let  ci^=c.    Then  xloga  =  logc.    Whence  a:  =  logc-f- loga. 

EXAMPLE 
Solve  for  x  8"^  =  324. 

Solution.  log  8-^  =  log  324, 

or  x  log  8  =  log  324. 

Whence  .  =  l2£^  =  1^  =  2.78+. 

logs         .9031 

The  student  must  overcome  his  hesitation  actually  to 
divide  one  logarithm  by  another  if,  as  here,  it  is  necessary. 


270  SECOND  COURSE  IN  ALGEBRA 

MISCELLANEOUS  EXERCISES 

1.  Can  you  find  the  logarithm  of  a  negative  number  to  a 
positive  base  ?    Explain. 

Find,  without  reference  to  the  table,  the  numerical  values  of; 

2.  log^O.  6.  51og^9. 

3.  log^8.  7.  log^8H-31og3  4. 

4.  log32.  8.  21og^81-41og3,27. 

5.  4  log^27.  9.  3  log^l25  +  2  log,25  -  2  log^o. 

10.  4  log,(i)  -  5  log/Jy)  +  2  log^T^- 
Simplify : 

11.  log  I  +  log  if  13.  log  -V-  +  log  U  -  log  i- 

12.  log ^V  -  log  f  |.  14.  2  log  3  +  3  log  2. 

15.  31og4  +  41og3-21og6. 
Show  that : 

16.  log  la  —  ^J=  log  (a  +  x)  +  log  (a  —  .r)  —  log  a. 

17.  log  V^'-x^  =  J  [log(«  +  x)-\-  log(«.  -  x)]. 

18.  log  Vr(.s'  -  a)  =  I  [log  s  +  log (.s  -a)y 

>  s  —  a 

lllogs  +  log(.s  -  h)  +  log(.v  -  c)  -  log{s  -  u)^ 

Solve,  using  logarithms  (obtain  results  to  four  figures) : 

20.  The  circumference  of  a  circle  is  2  wR.    (tt  =  3.1416, 
li  =  radius.) 

(a)  Eind  the  circumference  of  a  tdrcle  whose  radius  is  85    . 
inches. 

(b)  Eind  the  radius  of  a  circle  whose  circumference  is  3281 
centimeters. 


LOGARITHMS  271 

21.  The  area  of  a  circle  is  ttR^. 

(a)  Find  the  area  of  a  circle  whose  radius  is  5.672  feet. 

(b)  Find  the  radius  of  a  circle  whose  area  is  67.37  square 
feet. 

22.  The  area  of  the  surface  of  a  sphere  is  4  irR^. 

(a)  The  radius  of  the  earth  is  3958.79  miles.  Find  its 
surface. 

(b)  Find  the  length  of  the  equator. 

23.  The  volume  of  a  sphere  is  — 

o 

(a)  Find  the  radius  of  a  sphere  whose  volume  is  86  cubic  feet. 
(/>»)  Find  the  diameter  of  a  sphere  whose  volume  is  47  cubic 
inches. 

24.  If  the  hypotenuse  and  one  leg  of  a  right  triangle  are 
given,  the  other  leg  can  always  be  computed  by  logarithms. 

In  the  adjacent  figure  let  a  and  c  be  given  and  x  required. 
Then 

X  =  Vc^  -  a'  =  V{r-\-a)(c-a). 
Whence 
logic  =  i  log(c  +  «)  4-  1^  logO;  -  a). 

(a)  The  hypotenuse  of  a  right  triangle  is  377  and  one  leg 
is  288.    Find  the  other  leg. 

(b)  The  hypotenuse  of  a  right  triangle  is  1285  and  one  leg 
is  924.    Find  the  other  leg. 

25.  The  area  of  an  equilateral  triangle  whose  side  is  s  is 
—  V3.  Find  in  square  feet  the  area  of  an  equilateral  triangle 
whose  side  is  34.23  inches. 

Solve  for  x : 

26.  3^  =  25.  30.  3  =  (1.04)-^.  34.  10*^ "^  =  3. 

27.  64^  =  4.  31.  2^  =  64.  35.  8-^  +  '^  =  6. 

28.  16-^  =  1024.  32.  42*^+1  =  84.  36.  (.3)"^  =  5. 

29.  (-  2y  =  64.  33.  3^  +  ^  =  6561.  37.  (.07)-^  =  9. 


272  SECOND  COURSE  IN  ALGEBRA 

Find  the  number  of  digits  in : 

38.  (a)  3^2;    (b)  2^-,    (c)  2«  •  3« .  5'. 

39.  In  how  many  years  will  $1  double  itself  at  3%  interest 
compounded  annually  ?  * 

Solution.  At  the  end  of  one  year  the  amount  of  $1  at  3%  is  $1.03  ; 
at  the  end  of  two  years  it  is  |(1.03)  (1.03),  or  $(1.03)2 ;  at  the  end  of 
three  years  it  is  $(1.03)^,  and  at  the  end  of  x  years  it  is  $(1.03)^. 

If  X  is  the  number  of  years  required,  (1.03)^  =  2. 

Taking  the  logarithms  of  both  members  of  the  equation, 

X  log  1.03  =  log  2. 

o  1   •  log  2         .3010      ^^^  .  t^ 

Solvmg,  x  =  - — ^ = =  23.5  +  .  *& 

^  log  1.03      .0128  V^ 

40.  In  how  many  years  will  $1  double  itself  at  6%  interest 
compounded  annually? 

41.  In  how  many  years  will  any  sum  of  money  treble  itself 
at  4^  interest  compounded  annually  ? 

42.  In  how  many  years  will  $450  double  itself  at  3-|-% 
interest  compounded  annually  ? 

43.  In  how  many  years  will  $4000  amount  to  $7360.80  at 
5%  interest  compounded  annually  ? 

44.  About  300  years  ago  the  Dutch  paid  $24  for  the  island 
of  Manhattan.  At  4*^  compound  interest,  what  would  this 
payment  amount  to  at  the  present  time  ? 

45.  In  how  many  years  will  $12  double  itself  at  3% 
interest  compounded  semiannually  ? 

46.  Show  that  the  amount  of  P  dollars  in  t  years  at  ?•% 
interest  compounded  annually  is  P  (1  +  ry ;  compounded  semi- 
annually is  P(H--)  ;  compounded  quarterly  is  pfl  +  jj  ; 


and  compounded  monthly  is  P  f  1  +  — 

*  In  making  computations  of  this  nature  by  the  aid  of  logarithms,  care 
must  be  exercised  not  to  retain  more  significant  figures  in  the  jesult  than 
are  given  with  accuracy  by  the  process. 


LOGARITHMS  273 

47.  Find  the  amount  of  |5000  at  the  end  of  four  years, 
interest  at  4%  compounded  {(i)  annually;  (b)  semiannually; 
(c)  quarterly. 

48.  Find  the  amount  of  |4.12  at  the  end  of  five  years, 
interest  at  4^^,  compounded  quarterly. 

49.  Set  up  and  solve  the  equation  used  to  determine  the 
amount  which  should  be  paid  for  a  |5  certificate  to  be  paid 
in  five  years,  interest  at  4'^,  compounded  quarterly. 

Note.  It  is  not  a  little  remarkable  that  just  at  the  time  when 
Galileo  and  Kepler  were  turning  their  attention  to  the  laborious 
computation  of  the  orbits  of  planets,  Napier  should  be  devising  a 
method  which  simpUties  these  processes.  It  was  said  a  hundred 
years  ago,  before  astronomical  computations  became  so  complex  as 
they  now  are,  that  the  invention  of  logarithms,  by  shortening  the 
labors,  doubled  the  effective  life  of  the  astronomer.  To-day  the 
remark  is  well  inside  the  truth. 

In  the  presentation  of  the  subject  in  modern  textbooks  a  loga- 
rithm is  defined  as  an  exponent.  But  it  was  not  from  this  point  of 
view  that  they  were  first  considered  by  Napier.  In  fact  it  was  not 
till  long  after  his  time  that  the  theory  of  exponents  was  understood 
clearly  enough  to  admit  of  such  application.  This  relation  was 
noticed  by  the  mathematician  Euler,  about  one  hundred  and  fifty 
years  after  logarithms  were  invented. 

It  was  by  a  comparison  of  the  terms  of  certain  arithmetical  and 
geometrical  progressions  that  Napier  derived  his  logarithms.  They 
were  not  exactly  like  those  used  commonly  to-day,  for  the  base 
which  Napier  used  was  not  10.  Soon  after  the  publication  (1614) 
of  Napier's  work,  Henry  Briggs,  an  English  professor,  was  so  much 
impressed  with  its  importance  that  he  journeyed  to  Scotland  to  con- 
fer with  Napier  about  the  discovery.  It  is  probable  that  they  both 
saw  the  necessity  of  constructing  a  table  for  the  base  1 0,  and  to  this 
enormous  task  Briggs  applied  himself.  With  the  exception  of  one 
gap,  which  was  filled  in  by  another  computer,  Briggs's  tables  form 
the  basis  for  all  the  common  logarithms  which  have  appeared  from 
that  day  to  this. 

The  square  roots  and  the  cube  roots  on  the  following  page  are 
corrected  to  the  nearest  digit  in  the  third  decimal  place. 


274 


SECOND  COURSE  IN  ALGEBRA 


No. 

Squares 

Cubes 

Square 
Roots 

Cube 
Roots 

No. 

Squares 

Cubes 

Square 
Roots 

Cube 
Roots 

1 

1 

1 

1.000 

1.000 

51 

2,601 

132,651 

7.141 

3.708 

2 

4 

8 

1.414 

1.260 

52 

2,704 

140,608 

7.211 

3.733 

3 

9 

27 

1.7.32 

1.442 

53 

2,809 

148,877 

7.280 

3.756 

4 

16 

64 

2.000 

1.587 

54 

2,916 

157,464 

7.348 

3.780 

5 

25 

125 

2.236 

1.710 

55 

3,025 

166,375 

7.416 

3.803 

6 

36 

216 

2.449 

1.817 

56 

3,136 

175,616 

7.483 

3.826 

7 

49 

343 

2.646 

1.913 

57 

3,249 

185,193 

7.550 

3.849 

8 

64 

512 

2.828 

2.000 

58 

3,364 

195,112 

7.616 

3.871 

9 

81 

729 

3.000 

2.080 

59 

3,481 

205,379 

7.681 

3.893 

10 

100 

1,000 

3.162 

2.154 

60 

3,600 

216,000 

7.746 

3.915 

11 

121 

1,331 

3.317 

2.224 

61 

3,721 

226,981 

7.810 

3.936 

12 

144 

1,728 

3.464 

2.289 

62 

3,844 

238,328 

7.874 

3.958 

13 

169 

2,197 

3.606 

2.351 

63 

3,969 

250,047 

7.937 

3.979 

14 

196 

2,744 

3.742 

2.410 

64 

4,096 

262,144 

8.000 

4.000 

15 

225 

3,375 

3.873 

2.466 

65 

4,225 

274,625 

8.062 

4.021 

16 

256 

4,096 

4.000 

2.520 

66 

4,356 

287,496 

8.124 

4.041 

17 

289 

4,913 

4.123 

2.571 

67 

4,489 

300,763 

8.185 

4.062 

18 

324 

5,832 

4.243 

2.621 

68 

4,624 

314,4.32 

8.246 

4.082 

19 

361 

6,859 

4.359 

2.668 

69 

4,761 

328,509 

8.307 

4.102 

20 

400 

8,000 

4.472 

2.714 

70 

4,900 

^43,000 

8.367 

4.121 

21 

441 

9,261 

4.583 

2.759 

71 

5,041 

357,911 

8.426 

4.141. 

22 

484 

10,648 

4.690 

2.802 

72 

5,184 

373,248 

8.485 

4.160 

23 

529 

12,167 

4.796 

2.844 

73 

5,329 

389,017 

8.544 

4.179 

24 

576 

13,824 

4.899 

2.884 

74 

5,476 

405,224 

8.602 

4.198 

25 

625 

15,625 

6.000 

2.924 

75 

5,625 

421,875 

8.660 

4.217 

26 

676 

17,576 

5.099 

2.962 

76 

5,776 

438,976 

8.718 

4.236 

27 

729 

19,683 

5.196 

3.000 

77 

5,929 

.  456,533 

8.775 

4.254 

28 

784 

21,952 

5.292 

3.037 

78 

6,084 

474,552 

8.832 

4.273 

29 

841 

24,389 

5.385 

3.072 

79 

6,241 

493,039 

8.888 

4.291 

30 

900 

27,000 

5.477 

3.107 

80 

6,400 

512,000 

8.944 

4.309 

31 

961 

29,791 

5.568 

3.141 

81 

6,561 

531,441 

9.000 

4.327 

32 

1,024 

32,768 

5.657 

3.175 

8'.i 

6,724 

551,1368 

9.055 

4.344 

33 

1,089 

35,937 

5.745 

3.208 

83 

6,889 

571,787 

9.110 

4.362 

34 

1,156 

39,304 

5.831 

3.240 

84 

7,056 

592,704 

9.165 

4.380 

35 

1,225 

42,875 

5.916 

3.271 

85 

7,225 

614,125 

9.220 

4.397 

36 

1,296 

46,656 

6.000 

3.W2 

86 

7,396 

636,a56 

9.274 

4.414 

37 

1,369 

50,653 

6.083 

3.3;52 

87 

7,569 

6.58,503 

9.327 

4.431 

38 

1,444 

54,872 

6.164 

3.:362 

88 

7,744 

681,472 

9.^381 

4.448 

39 

1,521 

59,319 

6.245 

3.391 

89 

7,921 

704,969 

9.4M 

4.465 

40 

1,600 

64,000 

6.325 

3.420 

90 

8,100 

729,000 

9.487 

4.481 

41 

1,681 

68,921 

6.403 

3.448 

91 

8,281 

753,571 

9.539 

4.498 

42 

1,764 

74,088 

6.481 

3.476 

92 

8,464 

778,688 

9.592 

4.514 

43 

1,849 

79,.507 

6.557 

3.503 

93 

8,649 

804,357 

9.644 

4.531 

44 

1,936 

&5,184 

6.6;^3 

3.530 

M 

8,836 

830,584 

9.695 

4.547 

45 

2,025 

91,12,5 

6.708 

3.557 

95 

9,025 

857,375 

9.747 

4.563 

46 

2,116 

97,336 

6.782 

3.583 

96 

9,216 

884,7.% 

9.798 

4.579 

47 

2,209 

103,823 

6.856 

3.609 

97 

9,409 

912,673 

9.849 

4.595 

48 

2,304 

110,592 

6.928 

3.6;m 

98 

9,604 

941,192 

9.899 

4.610 

49 

2,401 

117,649 

7.000 

3.659 

99 

9,801 

970,299 

9.950 

4.626 

50 

2,500 

125,000 

7.071 

3.684 

100 

10,000 

1,000,000 

10.000 

4.642 

INDEX 


Abel,  224,  234 

Addition,  algebraic,  2  ;  commuta- 
tive law  of,  2  ;  of  fractions,  53 
Ahmes,  214 
Alternation,  238 
Antecedent,  235 
Antilogarithm,  260 
Apothem,  125 
Axes,  X-  and  y-,  69 
Axiom,  16 
Axioms,  16 

Base,  252,  257 

Binomial  theorem,  227 ;  extraction 

of  roots   by,   230 ;    rth  term  of 

(a  +  5)",  232-233 
Binomials,      difference      of      two 

squares,    28;    powers    of,    227; 

sum  or  difference  of  two  cubes, 

34 
Briggs,  253,  273 
Brouncker,  William,  60 

Cardan,  36,  165 
Cauchy,  224 
Characteristic,  255 
Checking,  rule  for,  19 
Circle,  179 

Complex  number,  159 
Conjugate  iniaginaries,  162 
Conjugate  real  radicals,  120 
Consequent,  235 
Constant,  244,  246,  248 
Coordinates  of  a  point,  70 
Cubes  of  numbers,  274 

Decimals,  repeating,  110 
Diophantos  of  Alexandria,  81 
Discriminant  of  a  quadratic  equa- 
tion, 172 
Distance,  x~  and  y-.  69 
Division,  by  logarithms,  263 ;  rule 
for,  7 


Elimination,  73 

Ellipse,  180 

Equation,  definition  of  an,  15  ;  of 
condition,  15  ;  derived,  77  ;  inde- 
pendent, 77;  root  of  an,  16 ;  satis- 
fying an,  16 

Equations,  definition  and  typical 
solution  of  irrational,  153 ;  equiv- 
alent, 17  ;  exponential,  269  ;  for- 
mation of,  with  given  roots,  168  ; 
homogeneous,  190 ;  with  imagi- 
nary roots,  136,  164  ;  indetermi- 
nate, 80 ;  involving  fractions, 
61 ;  linear,  solution  by  addition 
and  subtraction,  75 ;  in  one  un- 
known, rule  for  solving,  19 ; 
rule  for  the  solution  of  irrational, 
154 ;  in  several  unknowns,  80  ; 
solution  of,  by  factoring,  42 ; 
special  cases,  76  ;  use  of  division 
in,  196 

Euclid,  45,  123 

Euler,  234,  273 

Evolution,  by  logarithms,  265 ;  law 
of,  92 

Exponent,  logarithm  as  an,  253  ; 
meaning  of  a  fractional,  109 ; 
meaning  of  a  negative,  93 ;  mean- 
ing of  a  zero,  93 

Exponents,  fundamental  laws  of, 
91 

Expressions,  integral,  25;  reducible 
to  difference  of  two  squares,  33 

Extremes,  237 

Factor,  highest  common,  44 
Factor  Theorem,  36 
Factorial  notation,  232 
Factoring,  definition  of  the  process, 

25 ;   general  directions  for,  39 ; 

solution  of  equations  by,  42 
Factors,  prime,  25 
Fraction,  changes  of  sign  in  a,  49 


275 


276 


SECOND  COUESE  IN  ALGEBRA 


Fractions,  addition  and  subtrac- 
tion of,  53 ;  complex,  59 ;  division 
of,  57 ;  equivalent,  52  ;  multi- 
plication of,  56  ;  operations  on, 
47 

Function,  definition  of  a,  129; 
graph  of  a  cubic,  132 ;  graph  of  a 
linear,  130 ;  graph  of  a  quadratic, 
131 ;  notations  for  a,  130 

Functions,  129  ;  names  of,  129 

Galileo,  273 

Gauss,  134,  159,  165,  224 

Graph,  of  a  cubic  function,  132  ; 
of  a  linear  function,  130 ;  of  a 
quadratic  equation  in  two  vari- 
ables, 177 

Graphical  representation  of  nu- 
merical data,  184 

Graphical  solution  of  an  equation 
in  one  unknown,  134;  the  process 
of,  135 

Graphical  solution  of  a  linear  sys- 
tem, 69,  70 

Graphical  solution  of  a  quadratic 
system  in  two  variables,  181 

Hindu  mathematicians,  123 
Hyperbola,  178 

Identity,  15 

Imaginaries,  addition  and  subtrac- 
tion of,  159  ;  conjugate,  162  ;  defi- 
nitions, 158  ;  division  of,  162  ; 
equations  with  imaginary  roots, 
164 ;  factors  involving,  166 ;  mul- 
tiplication of,  160 ;  note  on  the 
use  of,  166 

Imaginary,  110 

Imaginary  roots,  136 

Index, 109 

Infinite  geometrical  series,  220 

Integral  expressions,  25 

Interpolation,  258 

Inversion,  238 

Involution,  by  logarithms,  264 ; 
law  of,  92 

Irrational  numbers,  109 

Kepler,  180,  273 

Klein,  Professor  Felix,  134 


La  Place,  224 

Leibnitz,  Gottfried  Wilhelm,  237 

Line,  straight,  130 

Logarithms,  definition  of,  252 ; 
antilogarithm,  260 ;  interpola- 
tion, 258  ;  systems  of,  253  ;  table 
of,  266-267 

Mantissa,  255,  263 

Means,  237;  arithmetical,  207;  geo- 
metrical, 217 

Multiple,  lowest  common,  50 

Multiplication,  law  of,  5  ;  rule  for, 
by  logarithms,  262 

Napier,  John  (Lord  of  Merchiston), 

234,  255,  273 
Newton,  137,  223,  234 
Numbers,    classification    of,    110; 

complex,  159;   imaginary,  110; 

irrational,  109;  orthotomic,  158; 

pure  imaginary,  158 ;    rational, 

109;  real,  110 

Operations,  order  of  fundamental,  1 
Origin,  70 
Oughtred,  237 

Parabola,  177 

Parentheses,  removal  of,  8 

Pascal's  triangle,  228 

Polynomials,  with  a  common 
monomial  factor,  26 ;  factored 
by  grouping  tenns  and  taking 
out  a  conmion  binomial  factor, 
27 

Powers,  sum  or  difference  of  two 
like,  37 

Probability  curve,  187 

Problems,  solution  of,  21 

Products,  important  special,  10 

Progressions,  connnon  difference 
of  arithmetical,  204  ;  definition 
of  arithmetical,  204;  definition  of 
geometrical,  214  ;  nth  term  of 
arithmetical,  205 ;  nth  term 
of  geometrical,  216 ;  ratio  of 
geometrical,  215 

Proportion,  236  ;  test  of  a,  237 

]*roportional,  fourth,  237;  mean, 
237;  third,  237 


INDEX 


277 


Proportions,  derived  by  addition, 
by  subtraction,  and  by  addition 
and  subtraction,  239 ;  from  equal 
products,  237 

Quadratic  equation,  character  of 
the  roots  of,  172  ;  comparison  of 
the  various  methods  of  solution 
of  the,  146 ;  formation  of,  with 
given  roots,  168 ;  number  of  roots 
of,  1 74 ;  relation  between  the  roots 
and  the  coefficients  of,  169 ;  solu- 
tion of,  by  completing  the  square, 
139 ;  solution  of,  by  formula,  144 

Quadratic  expressions,  factors  of, 
175 

Quadratic  trinomial,  30  ;  rule  for 
solving,  30  ;  the  general,  31;  rule 
for  solving  the  general,  82 

Quantity,  243 

lladicals,  109  ;  addition  and  sub- 
traction of,  116  ;  algebraic  sign 
of.  111;  conjugate,  120  ;  division 
of,  119  ;  factors  involving,  124 ; 
multiplication  of  real,  117;  simi- 
lar, 116;  simplification  of ,  112 

Radicand,  109 

Ratio,  235 

Rational  expressions,  25 

Rational  numbers,  109 

Rationalizing  factor,  119 

Remainder  Theorem,  35 

Root,  cube,  of  arithmetical  numbers, 
274 ;  cube,  of  monomials,  101 ;  defi- 
nition of  cube,  101 ;  definition  of 
square,  101;  principal,  101;  rule 
for  extracting  square,  106;  square, 
of  arithmetical  numbers,  106, 
274  ;  square,  of  a  monomial,  101 ; 
square,  of  polynomials,  103 ; 
square,  of  surd  expressions,  122 


Roots,  imaginary,  136  ;  imaginary, 
for  a  cubic  equation,  137;  table 
of  square  and  cube,  274 

Series,  arithmetical,  210  ;  geomet- 
rical, 219  ;  infinite  geometrical, 
220  ;  sum  of,  210 

Slide  rule,  269  • 

Squares  of  numbers,  274 

Stifel,  230 

Subtraction,  of  fractions,  53 ;  of 
polynomials,  3 

Surd,  110 

Systems,  determinate,  in  three 
variables,  82  ;  equivalent,  196 ; 
incompatible  or  inconsistent,  76 : 
simultaneous,  76 ;  solution  by 
addition  and  subtraction,  75 ;  so- 
lution by  substitution,  74;  solu- 
tion of,  when  one  equation  is 
linear  and  the  other  quadratic, 
188;  solution  of,  when  both  are 
quadratic,  191 ;  solution  of  a  lin- 
ear, in  two  variables,  by  graphs, 
69  ;  special  methods  for  solution 
of,  194 ;  symmetric,  193 

Table,  of  cubes  and  squares,  274 ; 

of  logarithms,  266-267 
Tartaglia,  36,  228 
Terms,  similar,  2 
Transposition,  17 
Trinomials,  perfect  squares,  28 

Variable,  130,  244 
Variation,  243;  direct,  243;  inverse, 
245 ;  joint,  247 

Wallis,  John,  60,  98,  237 

Zero  as  an  exponent,  meaning  of, 
93 


VB  35926 


459908 


UNIVERSITY  OF  CAUFORNIA  LIBRARY 


